What is the number of arrangements of 5 different things taken 3 at a time in which 2 particular thing always...
up vote
0
down vote
favorite
I know , there is formula to apply, s!(r-s+1)( n-s P k-s).
I tried with example,
n=5 ; r=3 and s=2.
I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.
But, if I solve manually with example,
consider n = {1,2,3,4,5}
then possible arrangements are as follows:
145
154
245
254
345
354
415
425
435
451
452
453
514
524
534
541
542
543.
Total count is 18.
Where am I going wrong
combinatorics permutations
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
add a comment |
up vote
0
down vote
favorite
I know , there is formula to apply, s!(r-s+1)( n-s P k-s).
I tried with example,
n=5 ; r=3 and s=2.
I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.
But, if I solve manually with example,
consider n = {1,2,3,4,5}
then possible arrangements are as follows:
145
154
245
254
345
354
415
425
435
451
452
453
514
524
534
541
542
543.
Total count is 18.
Where am I going wrong
combinatorics permutations
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
I'm referring this for formula: number 5 (b). books.google.co.in/…
– Ratnakar Chinchkar
Aug 20 at 16:32
2
A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product).
– JMoravitz
Aug 20 at 16:51
yea, thanks for the suggestion. Can you please help me out with this problem.
– Ratnakar Chinchkar
Aug 20 at 16:57
The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $binom{n-s}{r-s}cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed.
– JMoravitz
Aug 20 at 17:00
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 20 at 17:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know , there is formula to apply, s!(r-s+1)( n-s P k-s).
I tried with example,
n=5 ; r=3 and s=2.
I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.
But, if I solve manually with example,
consider n = {1,2,3,4,5}
then possible arrangements are as follows:
145
154
245
254
345
354
415
425
435
451
452
453
514
524
534
541
542
543.
Total count is 18.
Where am I going wrong
combinatorics permutations
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
I know , there is formula to apply, s!(r-s+1)( n-s P k-s).
I tried with example,
n=5 ; r=3 and s=2.
I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.
But, if I solve manually with example,
consider n = {1,2,3,4,5}
then possible arrangements are as follows:
145
154
245
254
345
354
415
425
435
451
452
453
514
524
534
541
542
543.
Total count is 18.
Where am I going wrong
combinatorics permutations
combinatorics permutations
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
edited Aug 20 at 17:12
N. F. Taussig
43.1k93254
43.1k93254
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
asked Aug 20 at 16:31
Ratnakar Chinchkar
11
11
I'm referring this for formula: number 5 (b). books.google.co.in/…
– Ratnakar Chinchkar
Aug 20 at 16:32
2
A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product).
– JMoravitz
Aug 20 at 16:51
yea, thanks for the suggestion. Can you please help me out with this problem.
– Ratnakar Chinchkar
Aug 20 at 16:57
The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $binom{n-s}{r-s}cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed.
– JMoravitz
Aug 20 at 17:00
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 20 at 17:15
add a comment |
I'm referring this for formula: number 5 (b). books.google.co.in/…
– Ratnakar Chinchkar
Aug 20 at 16:32
2
A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product).
– JMoravitz
Aug 20 at 16:51
yea, thanks for the suggestion. Can you please help me out with this problem.
– Ratnakar Chinchkar
Aug 20 at 16:57
The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $binom{n-s}{r-s}cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed.
– JMoravitz
Aug 20 at 17:00
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 20 at 17:15
I'm referring this for formula: number 5 (b). books.google.co.in/…
– Ratnakar Chinchkar
Aug 20 at 16:32
I'm referring this for formula: number 5 (b). books.google.co.in/…
– Ratnakar Chinchkar
Aug 20 at 16:32
2
2
A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product).
– JMoravitz
Aug 20 at 16:51
A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product).
– JMoravitz
Aug 20 at 16:51
yea, thanks for the suggestion. Can you please help me out with this problem.
– Ratnakar Chinchkar
Aug 20 at 16:57
yea, thanks for the suggestion. Can you please help me out with this problem.
– Ratnakar Chinchkar
Aug 20 at 16:57
The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $binom{n-s}{r-s}cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed.
– JMoravitz
Aug 20 at 17:00
The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $binom{n-s}{r-s}cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed.
– JMoravitz
Aug 20 at 17:00
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 20 at 17:15
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 20 at 17:15
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.
The formula that deals with what you want is $r!{n-s choose r-s}$. In this case, the number would be $3!{3 choose 1}=18$, exactly what you got.
answered Aug 20 at 16:44
SinTan1729
2,399622
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
add a comment |
up vote
0
down vote
I am not able to view your link, but here is how I would calculate it:
You are looking for your arrangement to contain both 4 and 5. So, choose one more number (from the remaining 3). Then permute the three numbers. My guess (and this is a guess) is that the formula you are looking at has something to do with multisets.
Anyway, choosing the third number, then permuting the three numbers would give you this calculation:
$$dbinom{3}{1}3! = 3cdot 3cdot 2cdot 1 = 18$$
which agrees with the number of arrangements you found by writing every one out.
answered Aug 20 at 16:39
InterstellarProbe
2,888727
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
add a comment |
up vote
-1
down vote
2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18
answered Nov 22 at 6:28
Megha
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2888968%2fwhat-is-the-number-of-arrangements-of-5-different-things-taken-3-at-a-time-in-wh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.
The formula that deals with what you want is $r!{n-s choose r-s}$. In this case, the number would be $3!{3 choose 1}=18$, exactly what you got.
answered Aug 20 at 16:44
SinTan1729
2,399622
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
add a comment |
up vote
2
down vote
The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.
The formula that deals with what you want is $r!{n-s choose r-s}$. In this case, the number would be $3!{3 choose 1}=18$, exactly what you got.
answered Aug 20 at 16:44
SinTan1729
2,399622
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
add a comment |
up vote
2
down vote
up vote
2
down vote
The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.
The formula that deals with what you want is $r!{n-s choose r-s}$. In this case, the number would be $3!{3 choose 1}=18$, exactly what you got.
answered Aug 20 at 16:44
SinTan1729
2,399622
The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.
The formula that deals with what you want is $r!{n-s choose r-s}$. In this case, the number would be $3!{3 choose 1}=18$, exactly what you got.
answered Aug 20 at 16:44
SinTan1729
2,399622
answered Aug 20 at 16:44
SinTan1729
2,399622
answered Aug 20 at 16:44
SinTan1729
2,399622
answered Aug 20 at 16:44
SinTan1729
2,399622
2,399622
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
add a comment |
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
You mean to say, 12 is when '4','5' are coming together?
– Ratnakar Chinchkar
Aug 20 at 17:01
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
@RatnakarChinchkar Yeah. I meant exactly that.
– SinTan1729
Aug 20 at 20:08
add a comment |
up vote
0
down vote
I am not able to view your link, but here is how I would calculate it:
You are looking for your arrangement to contain both 4 and 5. So, choose one more number (from the remaining 3). Then permute the three numbers. My guess (and this is a guess) is that the formula you are looking at has something to do with multisets.
Anyway, choosing the third number, then permuting the three numbers would give you this calculation:
$$dbinom{3}{1}3! = 3cdot 3cdot 2cdot 1 = 18$$
which agrees with the number of arrangements you found by writing every one out.
answered Aug 20 at 16:39
InterstellarProbe
2,888727
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
add a comment |
up vote
0
down vote
I am not able to view your link, but here is how I would calculate it:
You are looking for your arrangement to contain both 4 and 5. So, choose one more number (from the remaining 3). Then permute the three numbers. My guess (and this is a guess) is that the formula you are looking at has something to do with multisets.
Anyway, choosing the third number, then permuting the three numbers would give you this calculation:
$$dbinom{3}{1}3! = 3cdot 3cdot 2cdot 1 = 18$$
which agrees with the number of arrangements you found by writing every one out.
answered Aug 20 at 16:39
InterstellarProbe
2,888727
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
add a comment |
up vote
0
down vote
up vote
0
down vote
I am not able to view your link, but here is how I would calculate it:
You are looking for your arrangement to contain both 4 and 5. So, choose one more number (from the remaining 3). Then permute the three numbers. My guess (and this is a guess) is that the formula you are looking at has something to do with multisets.
Anyway, choosing the third number, then permuting the three numbers would give you this calculation:
$$dbinom{3}{1}3! = 3cdot 3cdot 2cdot 1 = 18$$
which agrees with the number of arrangements you found by writing every one out.
answered Aug 20 at 16:39
InterstellarProbe
2,888727
I am not able to view your link, but here is how I would calculate it:
You are looking for your arrangement to contain both 4 and 5. So, choose one more number (from the remaining 3). Then permute the three numbers. My guess (and this is a guess) is that the formula you are looking at has something to do with multisets.
Anyway, choosing the third number, then permuting the three numbers would give you this calculation:
$$dbinom{3}{1}3! = 3cdot 3cdot 2cdot 1 = 18$$
which agrees with the number of arrangements you found by writing every one out.
answered Aug 20 at 16:39
InterstellarProbe
2,888727
answered Aug 20 at 16:39
InterstellarProbe
2,888727
answered Aug 20 at 16:39
InterstellarProbe
2,888727
answered Aug 20 at 16:39
InterstellarProbe
2,888727
2,888727
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
add a comment |
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3.
– Ratnakar Chinchkar
Aug 20 at 16:54
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
@Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$begin{matrix}145 \ 154 \ 245 \ 254 \ 345 \ 354 \ 451 \ 541 \ 452 \ 542 \ 453 \ 543end{matrix}$$
– InterstellarProbe
Aug 20 at 17:35
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
Yea, thanks a lot.
– Ratnakar Chinchkar
Aug 20 at 17:43
add a comment |
up vote
-1
down vote
2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18
answered Nov 22 at 6:28
Megha
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
add a comment |
up vote
-1
down vote
2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18
answered Nov 22 at 6:28
Megha
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
add a comment |
up vote
-1
down vote
up vote
-1
down vote
2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18
answered Nov 22 at 6:28
Megha
1
2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18
answered Nov 22 at 6:28
Megha
1
answered Nov 22 at 6:28
Megha
1
answered Nov 22 at 6:28
Megha
1
answered Nov 22 at 6:28
Megha
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 22 at 7:06
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2888968%2fwhat-is-the-number-of-arrangements-of-5-different-things-taken-3-at-a-time-in-wh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I'm referring this for formula: number 5 (b). books.google.co.in/…
– Ratnakar Chinchkar
Aug 20 at 16:32
2
A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product).
– JMoravitz
Aug 20 at 16:51
yea, thanks for the suggestion. Can you please help me out with this problem.
– Ratnakar Chinchkar
Aug 20 at 16:57
The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $binom{n-s}{r-s}cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed.
– JMoravitz
Aug 20 at 17:00
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Aug 20 at 17:15