Permutation in equation!

Probability Level 3

Positive integers n 1 < n 2 < n 3 < n 4 < n 5 n_1<n_2<n_3<n_4<n_5 are such that n 1 + n 2 + n 3 + n 4 + n 5 = 20. n_1+n_2+n_3+n_4+n_5=20. Find the number of distinct tuples of ( n 1 , n 2 , n 3 , n 4 , n 5 ) \left(n_1,n_2,n_3,n_4,n_5\right) .


The answer is 7.

Rajdeep Brahma by rajdeep brahma , 21, India

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2 solutions

Aman Dubey
Apr 5, 2017

as minimum value of n (i) can be i,

let n(i) = i+a (i). where all ai's >=0. so we get

a1+a2+a3+a4+a5=5

for any partition of 5 (eg. 5=0+0+0+0+5) there corresponds only one solution to above eqn. hence number of solution is same as number partitions of 5, which is 7.

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Rajdeep Brahma
Mar 28, 2017

Set n2=n1+t1,n3=n2+t2,n4=n3+t3,n5=n4+t4 where all ti's>0. You get 5n1+4t1+3t2+2t3+t4=0. It's trivial to check that it has 7 solutions in positive integers.

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