Let n_1

Question

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) is

Accepted Answer

Correct answer: 7. (7) n2 = n1 + t1 + 1 n3 = n2 + t2 + 1 n4 = n3 + t3 + 1 n5 = n4 + t4 + 1 The given equation becomes 5n1 + 4t1 + 3t2 + 2t3 + t4 = 10 where n1 ≥ 1 ; t1 ≥ 0 n1 = t0 + 1 ⇒ 5t0 + 4t1 + 3t2 + 2t3 + t4 = 5 t0 = 1 will yield only 1 solution. so t0 = 0, 4t1 + 3t2 + 2t3 + t4 = 5. t1 = 0 = t2. there will be 3 solution t1 = 0, t2 = 1 will yield 2 solution. t1 = 1, t2 must be zero 1 solution. Hence in total there will be 7 solution.

$Let n_{1} < n_{2} < n_{3} < n_{4} < n_{5} be positive integers such that n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20. Then the number of such distinct arrangements (n_{1}, n_{2}, n_{3}, n_{4}, n_{5}) is$

Solution

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