# Candy

How many ways can divide 10 identical candies to 5 children?

Result

x =  2002

#### Solution:

$n = 10 \ \\ k = 5 \ \\ \ \\ C_{{ 5}}(14) = \dbinom{ 14}{ 5} = \dfrac{ 14! }{ 5!(14-5)!} = \dfrac{ 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 } { 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } = 2002 \ \\ \ \\ x = { { n+k-1 } \choose k } = { { 10+5-1 } \choose 5 } = 2002$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Math student
This should be 2002 not 1001 Zak
yes, combinations with repetition Math student
Isn't it 14 choose 9..?
And I also think it's group distribution theory and not combinations precisely as n<r. Math student
Which is 1001. Math student
14C9 is 2002. Tips to related online calculators
Would you like to compute count of combinations?

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