# how many ways are there to

• Nov 3rd 2008, 10:19 AM
pandakrap
how many ways are there to
You have 10(indistinguishable) bananas and 10(indistinguishable) strawberries.

a) how many ways are there to pass all 20 pieces of furit out to 10 children?
b) how many ways are there to pass all 20 pieces of fruit out to 10 children so that each child gets 2 pieces?
• Nov 3rd 2008, 10:30 AM
Plato
Quote:

Originally Posted by pandakrap
You have 10(indistinguishable) bananas and 10(indistinguishable) strawberries.
a) how many ways are there to pass all 20 pieces of furit out to 10 children?
b) how many ways are there to pass all 20 pieces of fruit out to 10 children so that each child gets 2 pieces?

a) ${{10+10-1} \choose {10}}^2$

Can you explain this?
• Nov 3rd 2008, 11:22 AM
pandakrap
i'm sorry, bit confused. could you explain that a little. thanks :)
• Nov 3rd 2008, 11:33 AM
Plato
Quote:

Originally Posted by pandakrap
could you explain that a little.

No, that is what I asked you to do.
Tell us where any of those formulas could have come from.
• Nov 3rd 2008, 11:53 AM
pandakrap
well i get the first one, 10 combinations of the banana with repetition, multiplied by 10 combinations of the strawberry with repetition,

but the second one.. i'm bit confused.
what you did is 20 choose 10, but... does that guarentee 2 fruits per child?
• Nov 3rd 2008, 01:06 PM
Plato
Quote:

Originally Posted by pandakrap
well i get the first one, 10 combinations of the banana with repetition, multiplied by 10 combinations of the strawberry with repetition,

but the second one.. i'm bit confused.
what you did is 20 choose 10, but... does that guarentee 2 fruits per child?

That is good on #1.

P.S. I apologize for give you a gross over count before on #2.
We can have anywhere from 0 to 5 selections of 2 B’s.

But the interesting idea here is that if we have 3 selections of 2B’s there must be 3 selections of 2S’s and there must be 4 mixed selections, [B,S].
So in this particular case we can assign these in $\frac {10!} {[3!]^2[4!]}$ ways.

Thus the general solution is: $\sum\limits_{k = 0}^5 {\frac{{10!}}
{{\left[ {\left( {5 - k} \right)!} \right]^2 \left[ {\left( {2k} \right)!} \right]}}} = {\text{8953}}$
.
• Nov 3rd 2008, 01:59 PM
pandakrap
thanks for the help, i get it,