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Let n1 le n2< n3 < n4< n5 be positive ...

Let `n_1 le n_2< n_3 < n_4< n_5` be positive integers such that `n_1+n_2+n_3+n_4+n_5=20.` then the number of such distinct arrangements `n_1, n_2, n_3, n_4, n_5` is

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The correct Answer is:
7
If `n_(1),n_(2),n_(3),n_(4)` take minimum values 1,2,3,4 respectively, then `n_(5)` will be maximum 10.
`therefore`corresponding to `n_(5)=10` there is only one solution
`n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=4`
Corresponnding to `n_(5)=9`, we can have,
`n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=5` i.e., one solution
Corresponding to `n_(5)=8`, we have,
`n_(1)=1,n_(2)=2,n_(3)=3,n_(4)=6`
or `n_(1)=1,n_(2)=2,n_(3)=4,n_(4)=5` i.e., two solutions
corresponding to `n_(5)=5`, we can have
`n_(1)=1,n_(2)=2,n_(3)=4, n_(4)=6`
or `n_(1)=1,n_(2)=3,n_(3)=4,n_(4)=5` i.e., two solutions
Corresponding to `n_(5)=6` we can have
`n_(1)=2,n_(2)=3,n_(3)=4,n_(4)=5` i.e., one solution
thus, there can be 7 solutions.
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