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← JEE Advanced 2014 · Paper 1
MathematicsInteger+3Question 53 of 60

JEE Advanced 2014 · Paper 1, Mathematics question 53

Let n1<n2<n3<n4<n5n_1<n_2<n_3<n_4<n_5 be positive integers such that n1+n2+n3+n4+n5=20n_1+n_2+n_3+n_4+n_5=20. Then the number of such distinct arrangements (n1,n2,n3,n4,n5)(n_1,n_2,n_3,n_4,n_5) is

Answer and worked solution

Official answer: 7

Shift. Put mi=ni−im_i=n_i-i. Then 0≤m1≤m2≤⋯≤m50\le m_1\le m_2\le\cdots\le m_5 and
∑mi=20−(1+2+3+4+5)=5.\sum m_i=20-(1+2+3+4+5)=5.

Count. Non-decreasing sequences of five non-negative integers with sum 5 correspond to partitions of 5 into at most 5 parts:
5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1.5,\ 4+1,\ 3+2,\ 3+1+1,\ 2+2+1,\ 2+1+1+1,\ 1+1+1+1+1.
That is 7.

Answer: 7

Topics:Distribution and arrangements
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