In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at...
- In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?
I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below
$$binom{6}{2}$$
This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that
$$binom{4}{1}binom{3}{1}$$
We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.
$$binom{6}{2}binom{4}{2}binom{2}{2}$$
Finally, I'm combining all these results and we have that
$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$
I've gone wrong somewhere. Could you assist me with that?
combinations
add a comment |
- In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?
I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below
$$binom{6}{2}$$
This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that
$$binom{4}{1}binom{3}{1}$$
We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.
$$binom{6}{2}binom{4}{2}binom{2}{2}$$
Finally, I'm combining all these results and we have that
$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$
I've gone wrong somewhere. Could you assist me with that?
combinations
1
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
– Shubham Johri
Nov 27 '18 at 19:27
add a comment |
- In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?
I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below
$$binom{6}{2}$$
This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that
$$binom{4}{1}binom{3}{1}$$
We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.
$$binom{6}{2}binom{4}{2}binom{2}{2}$$
Finally, I'm combining all these results and we have that
$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$
I've gone wrong somewhere. Could you assist me with that?
combinations
- In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?
I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below
$$binom{6}{2}$$
This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that
$$binom{4}{1}binom{3}{1}$$
We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.
$$binom{6}{2}binom{4}{2}binom{2}{2}$$
Finally, I'm combining all these results and we have that
$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$
I've gone wrong somewhere. Could you assist me with that?
combinations
combinations
asked Nov 27 '18 at 19:18
Enzo
996
996
1
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
– Shubham Johri
Nov 27 '18 at 19:27
add a comment |
1
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
– Shubham Johri
Nov 27 '18 at 19:27
1
1
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
– Shubham Johri
Nov 27 '18 at 19:27
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
– Shubham Johri
Nov 27 '18 at 19:27
add a comment |
2 Answers
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Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.
If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
$$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$
add a comment |
John get 2, that is ${6choose 2}$.
Now from the rest of 4 Carl gets:
$1$ toy, that is ${4choose 1}$ way or
$3$ toys, that is ${4choose 3}$ ways or
$2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.
add a comment |
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2 Answers
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2 Answers
2
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oldest
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Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.
If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
$$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$
add a comment |
Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.
If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
$$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$
add a comment |
Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.
If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
$$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$
Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.
If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
$$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$
answered Nov 27 '18 at 19:28
N. F. Taussig
43.5k93355
43.5k93355
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John get 2, that is ${6choose 2}$.
Now from the rest of 4 Carl gets:
$1$ toy, that is ${4choose 1}$ way or
$3$ toys, that is ${4choose 3}$ ways or
$2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.
add a comment |
John get 2, that is ${6choose 2}$.
Now from the rest of 4 Carl gets:
$1$ toy, that is ${4choose 1}$ way or
$3$ toys, that is ${4choose 3}$ ways or
$2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.
add a comment |
John get 2, that is ${6choose 2}$.
Now from the rest of 4 Carl gets:
$1$ toy, that is ${4choose 1}$ way or
$3$ toys, that is ${4choose 3}$ ways or
$2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.
John get 2, that is ${6choose 2}$.
Now from the rest of 4 Carl gets:
$1$ toy, that is ${4choose 1}$ way or
$3$ toys, that is ${4choose 3}$ ways or
$2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.
answered Nov 27 '18 at 19:27
greedoid
37.9k114794
37.9k114794
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Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
– Shubham Johri
Nov 27 '18 at 19:27