How many selections of three cards can be made from cards bearing the letters of the word EXAMINATION if …?
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$11$ cards each bear a letter, and together they can be made to spell the word "EXAMINATION". $3$ Cards are selected from the $11$ cards and the order of selection is not important. Find how many selections can be made if $2$ cards of the $3$ cards bear the same letter.
My approach.
I grouped the same letter together in a group and single letter is another group.
E X A M I N T O
AA II NN
This shows I got $8$ separate groups of different letters.
Now I understand I need cards from A, I or N first. So I need to choose $1$ from $3$ groups with the same letter. $to~_3C_1$
Now I am left with one more card I need to pick. This means I can't pick any of the $2$ groups with same letter remaining. So total sets I can only choose from now is deducted by $3~to~5~to~_5C_1$
So the answer is $to~_3C_1 cdot _5C_1$
Why is this wrong?
The workbook answer shows $_3C_1 cdot _7C_1 $
For the second one, if I pick one card out of the remaining $7$ groups. There is a chance I may pick the group with $2$ same letters, meaning the total cards I have now is $4$ instead of the required $3$.
combinatorics combinations
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
asked Nov 24 at 4:53
mutu mumu
374
add a comment |
up vote
2
down vote
favorite
$11$ cards each bear a letter, and together they can be made to spell the word "EXAMINATION". $3$ Cards are selected from the $11$ cards and the order of selection is not important. Find how many selections can be made if $2$ cards of the $3$ cards bear the same letter.
My approach.
I grouped the same letter together in a group and single letter is another group.
E X A M I N T O
AA II NN
This shows I got $8$ separate groups of different letters.
Now I understand I need cards from A, I or N first. So I need to choose $1$ from $3$ groups with the same letter. $to~_3C_1$
Now I am left with one more card I need to pick. This means I can't pick any of the $2$ groups with same letter remaining. So total sets I can only choose from now is deducted by $3~to~5~to~_5C_1$
So the answer is $to~_3C_1 cdot _5C_1$
Why is this wrong?
The workbook answer shows $_3C_1 cdot _7C_1 $
For the second one, if I pick one card out of the remaining $7$ groups. There is a chance I may pick the group with $2$ same letters, meaning the total cards I have now is $4$ instead of the required $3$.
combinatorics combinations
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
asked Nov 24 at 4:53
mutu mumu
374
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$11$ cards each bear a letter, and together they can be made to spell the word "EXAMINATION". $3$ Cards are selected from the $11$ cards and the order of selection is not important. Find how many selections can be made if $2$ cards of the $3$ cards bear the same letter.
My approach.
I grouped the same letter together in a group and single letter is another group.
E X A M I N T O
AA II NN
This shows I got $8$ separate groups of different letters.
Now I understand I need cards from A, I or N first. So I need to choose $1$ from $3$ groups with the same letter. $to~_3C_1$
Now I am left with one more card I need to pick. This means I can't pick any of the $2$ groups with same letter remaining. So total sets I can only choose from now is deducted by $3~to~5~to~_5C_1$
So the answer is $to~_3C_1 cdot _5C_1$
Why is this wrong?
The workbook answer shows $_3C_1 cdot _7C_1 $
For the second one, if I pick one card out of the remaining $7$ groups. There is a chance I may pick the group with $2$ same letters, meaning the total cards I have now is $4$ instead of the required $3$.
combinatorics combinations
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
asked Nov 24 at 4:53
mutu mumu
374
$11$ cards each bear a letter, and together they can be made to spell the word "EXAMINATION". $3$ Cards are selected from the $11$ cards and the order of selection is not important. Find how many selections can be made if $2$ cards of the $3$ cards bear the same letter.
My approach.
I grouped the same letter together in a group and single letter is another group.
E X A M I N T O
AA II NN
This shows I got $8$ separate groups of different letters.
Now I understand I need cards from A, I or N first. So I need to choose $1$ from $3$ groups with the same letter. $to~_3C_1$
Now I am left with one more card I need to pick. This means I can't pick any of the $2$ groups with same letter remaining. So total sets I can only choose from now is deducted by $3~to~5~to~_5C_1$
So the answer is $to~_3C_1 cdot _5C_1$
Why is this wrong?
The workbook answer shows $_3C_1 cdot _7C_1 $
For the second one, if I pick one card out of the remaining $7$ groups. There is a chance I may pick the group with $2$ same letters, meaning the total cards I have now is $4$ instead of the required $3$.
combinatorics combinations
combinatorics combinations
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
asked Nov 24 at 4:53
mutu mumu
374
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
asked Nov 24 at 4:53
mutu mumu
374
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
edited Nov 24 at 10:59
N. F. Taussig
43.4k93355
43.4k93355
asked Nov 24 at 4:53
mutu mumu
374
asked Nov 24 at 4:53
mutu mumu
374
asked Nov 24 at 4:53
mutu mumu
374
374
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The first part of your answer is right. You have 3 ways to choose from $A, I, N$. However, after that you're only considering picking from a group of letters which occur only once, in this case $E, X, M, T, O$. This would lead you to miss a case like $AAN$ where the second letter could also come from one of the repeated ones. So after picking one letter from $3$ possible options with $3choose1$ number of ways, you pick one letter from the rest of all the remaining distinct letters ($n-1 = 8-1 =7$) as there are $8$ distinct letters and one letter has already been chosen. You can choose $1$ from them in $7choose1$ ways, giving you the answer
$${3choose1}{7choose1}$$
answered Nov 24 at 5:45
Sauhard Sharma
52611
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The first part of your answer is right. You have 3 ways to choose from $A, I, N$. However, after that you're only considering picking from a group of letters which occur only once, in this case $E, X, M, T, O$. This would lead you to miss a case like $AAN$ where the second letter could also come from one of the repeated ones. So after picking one letter from $3$ possible options with $3choose1$ number of ways, you pick one letter from the rest of all the remaining distinct letters ($n-1 = 8-1 =7$) as there are $8$ distinct letters and one letter has already been chosen. You can choose $1$ from them in $7choose1$ ways, giving you the answer
$${3choose1}{7choose1}$$
answered Nov 24 at 5:45
Sauhard Sharma
52611
add a comment |
up vote
1
down vote
accepted
The first part of your answer is right. You have 3 ways to choose from $A, I, N$. However, after that you're only considering picking from a group of letters which occur only once, in this case $E, X, M, T, O$. This would lead you to miss a case like $AAN$ where the second letter could also come from one of the repeated ones. So after picking one letter from $3$ possible options with $3choose1$ number of ways, you pick one letter from the rest of all the remaining distinct letters ($n-1 = 8-1 =7$) as there are $8$ distinct letters and one letter has already been chosen. You can choose $1$ from them in $7choose1$ ways, giving you the answer
$${3choose1}{7choose1}$$
answered Nov 24 at 5:45
Sauhard Sharma
52611
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The first part of your answer is right. You have 3 ways to choose from $A, I, N$. However, after that you're only considering picking from a group of letters which occur only once, in this case $E, X, M, T, O$. This would lead you to miss a case like $AAN$ where the second letter could also come from one of the repeated ones. So after picking one letter from $3$ possible options with $3choose1$ number of ways, you pick one letter from the rest of all the remaining distinct letters ($n-1 = 8-1 =7$) as there are $8$ distinct letters and one letter has already been chosen. You can choose $1$ from them in $7choose1$ ways, giving you the answer
$${3choose1}{7choose1}$$
answered Nov 24 at 5:45
Sauhard Sharma
52611
The first part of your answer is right. You have 3 ways to choose from $A, I, N$. However, after that you're only considering picking from a group of letters which occur only once, in this case $E, X, M, T, O$. This would lead you to miss a case like $AAN$ where the second letter could also come from one of the repeated ones. So after picking one letter from $3$ possible options with $3choose1$ number of ways, you pick one letter from the rest of all the remaining distinct letters ($n-1 = 8-1 =7$) as there are $8$ distinct letters and one letter has already been chosen. You can choose $1$ from them in $7choose1$ ways, giving you the answer
$${3choose1}{7choose1}$$
answered Nov 24 at 5:45
Sauhard Sharma
52611
answered Nov 24 at 5:45
Sauhard Sharma
52611
answered Nov 24 at 5:45
Sauhard Sharma
52611
answered Nov 24 at 5:45
Sauhard Sharma
52611
52611
add a comment |
add a comment |
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