Candy

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

Result

x =  2002

Solution:

C5(14)=(145)=14!5!(145)!=141312111054321=2002 n=10 k=5  x=(n+k1k)=(10+515)=2002C_{{ 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 \ \\ n=10 \ \\ k=5 \ \\ \ \\ x={ { n+k-1 } \choose k }={ { 10+5-1 } \choose 5 }=2002



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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Showing 5 comments:
#
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.

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