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Question

Let n1<n2<n3<n4<n5 be positive integers such that n1+n2+n3+n4+n5=20. Then the number of such distinct arrangements (n1,n2,n3,n4,n5) is ___

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Solution

n1,n2,n3,n4 and n5are positive integers such that n1<n2<n3<n4<n5
then for n1+n2+n3+n4+n5=20
If n1,n2,n3,n4, take minimum values 1,2,3,4 respectively then n5=10, there is only one soution n1=1,n2=2,n3=3,n4=5 i.e., one solution Corresponding to n5=9,we can have, only soution n1=1,n2=n3=3,n4=4.
Corresponding to n5=8, we can have, only solution
n1=1,n2=2,n3=3,n4=6
or n1=1,n2=2,n3=4,n4,=4
i.e., two solutions.
For n5=7, we can have n1=1,n2=1n3=4,n4=5 i.e. 2 solutions
For n5=6, we can have
n1=2,n2=3,n3=4,n4=5
i.e., one solution
Thus there can be 7 solutions.

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