Mixing
Combinatorics Gifts Question
$begingroup$
1) How many ways can 7 identical gifts be distributed among 10 children if no child is allowed to get more
than 1 gift?
2) How many ways can 7 identical gifts be distributed among 10 children if children are allowed to get more
than 1 gift?
For problem 1, I was able to find an answer, but I am unsure if it is correct. The way I did this was:
Number of ways of dividing n identical gifts to r children is:
n+r-1 choose r-1, where n=7 and r=3, then 16C9 or
16!/(7!*9!) ways.
For problem 2, I am unsure of where to start?
probability combinatorics
asked Jan 31 at 21:14
Taffies1Taffies1
52
$endgroup$
add a comment |
$begingroup$
1) How many ways can 7 identical gifts be distributed among 10 children if no child is allowed to get more
than 1 gift?
2) How many ways can 7 identical gifts be distributed among 10 children if children are allowed to get more
than 1 gift?
For problem 1, I was able to find an answer, but I am unsure if it is correct. The way I did this was:
Number of ways of dividing n identical gifts to r children is:
n+r-1 choose r-1, where n=7 and r=3, then 16C9 or
16!/(7!*9!) ways.
For problem 2, I am unsure of where to start?
probability combinatorics
asked Jan 31 at 21:14
Taffies1Taffies1
52
$endgroup$
2
$begingroup$
For $1$, you just need to pick the seven children who will get a gift. The calculation you did is more appropriate for $2$.
$endgroup$
– lulu
Jan 31 at 21:17
add a comment |
$begingroup$
1) How many ways can 7 identical gifts be distributed among 10 children if no child is allowed to get more
than 1 gift?
2) How many ways can 7 identical gifts be distributed among 10 children if children are allowed to get more
than 1 gift?
For problem 1, I was able to find an answer, but I am unsure if it is correct. The way I did this was:
Number of ways of dividing n identical gifts to r children is:
n+r-1 choose r-1, where n=7 and r=3, then 16C9 or
16!/(7!*9!) ways.
For problem 2, I am unsure of where to start?
probability combinatorics
asked Jan 31 at 21:14
Taffies1Taffies1
52
$endgroup$
1) How many ways can 7 identical gifts be distributed among 10 children if no child is allowed to get more
than 1 gift?
2) How many ways can 7 identical gifts be distributed among 10 children if children are allowed to get more
than 1 gift?
For problem 1, I was able to find an answer, but I am unsure if it is correct. The way I did this was:
Number of ways of dividing n identical gifts to r children is:
n+r-1 choose r-1, where n=7 and r=3, then 16C9 or
16!/(7!*9!) ways.
For problem 2, I am unsure of where to start?
probability combinatorics
probability combinatorics
asked Jan 31 at 21:14
Taffies1Taffies1
52
asked Jan 31 at 21:14
Taffies1Taffies1
52
asked Jan 31 at 21:14
Taffies1Taffies1
52
asked Jan 31 at 21:14
Taffies1Taffies1
52
asked Jan 31 at 21:14
Taffies1Taffies1
52
52
2
$begingroup$
For $1$, you just need to pick the seven children who will get a gift. The calculation you did is more appropriate for $2$.
$endgroup$
– lulu
Jan 31 at 21:17
add a comment |
2
$begingroup$
For $1$, you just need to pick the seven children who will get a gift. The calculation you did is more appropriate for $2$.
$endgroup$
– lulu
Jan 31 at 21:17
2
2
$begingroup$
For $1$, you just need to pick the seven children who will get a gift. The calculation you did is more appropriate for $2$.
$endgroup$
– lulu
Jan 31 at 21:17
$begingroup$
For $1$, you just need to pick the seven children who will get a gift. The calculation you did is more appropriate for $2$.
$endgroup$
– lulu
Jan 31 at 21:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For 1, you just choose the seven children who will get gifts, so $10choose 7$ ways
Your answer to 1 is the correct approach for 2
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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1 Answer
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$begingroup$
For 1, you just choose the seven children who will get gifts, so $10choose 7$ ways
Your answer to 1 is the correct approach for 2
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
For 1, you just choose the seven children who will get gifts, so $10choose 7$ ways
Your answer to 1 is the correct approach for 2
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
For 1, you just choose the seven children who will get gifts, so $10choose 7$ ways
Your answer to 1 is the correct approach for 2
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
$endgroup$
For 1, you just choose the seven children who will get gifts, so $10choose 7$ ways
Your answer to 1 is the correct approach for 2
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 21:18
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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2
$begingroup$
For $1$, you just need to pick the seven children who will get a gift. The calculation you did is more appropriate for $2$.
$endgroup$
– lulu
Jan 31 at 21:17