JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Permutations and Combinations

Let

n_{1} < 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})

Answer: 7

Solution

We need to find the number of strictly increasing 5-tuples of positive integers that sum to 20. Since

n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

and all are positive, the minimum sum is

1 + 2 + 3 + 4 + 5 = 15

. We have 5 units to distribute among the gaps. By systematic enumeration:

{1, 2, 3, 4, 10}, {1, 2, 3, 5, 9}, {1, 2, 3, 6, 8}, {1, 2, 4, 5, 8}, {1, 2, 4, 6, 7}, {1, 3, 4, 5, 7}, {2, 3, 4, 5, 6}

. There are 7 such arrangements.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

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