Let n_1

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(D) We are given $n_1 < n_2 < n_3 < n_4 < n_5$ where $n_i \in \mathbb{Z}^+$ and $\sum_{i=1}^5 n_i = 20$.Let $n_1 = a_1$,$n_2 = a_1 + a_2 + 1$,$n_3 = a

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) We are given $n_1 Let $n_1 = a_1$,$n_2 = a_1 + a_2 + 1$,$n_3 = a_1 + a_2 + a_3 + 2$,$n_4 = a_1 + a_2 + a_3 + a_4 + 3$,$n_5 = a_1 + a_2 + a_3 + a_4 + a_5 + 4$,where $a_1 \geq 1$ and $a_2, a_3, a_4, a_5 \geq 0$.Substituting these into the sum:$5a_1 + 4a_2 + 3a_3 + 2a_4 + a_5 + 10 = 20 \implies 5a_1 + 4a_2 + 3a_3 + 2a_4 + a_5 = 10$.Since $a_1 \geq 1$,let $a_1 = 1 + k$ where $k \geq 0$. Then $5(1+k) + 4a_2 + 3a_3 + 2a_4 + a_5 = 10 \implies 5k + 4a_2 + 3a_3 + 2a_4 + a_5 = 5$.If $k=1$,then $a_2=a_3=a_4=a_5=0$,which gives $(n_1, n_2, n_3, n_4, n_5) = (2, 3, 4, 5, 6)$.If $k=0$,then $4a_2 + 3a_3 + 2a_4 + a_5 = 5$. The possible non-negative integer solutions $(a_2, a_3, a_4, a_5)$ are:$1$) $(1, 0, 0, 1) \implies (1, 3, 4, 5, 7)$$2$) $(1, 0, 1, -1)$ (Invalid)$3$) $(0, 1, 1, 0) \implies (1, 2, 4, 6, 7)$$4$) $(0, 1, 0, 2) \implies (1, 2, 3, 6, 8)$$5$) $(0, 0, 2, 1) \implies (1, 2, 3, 5, 9)$$6$) $(0, 0, 1, 3) \implies (1, 2, 3, 4, 10)$$7$) $(0, 0, 0, 5) \implies (1, 2, 3, 4, 10)$ (Wait,checking manually: $(1, 2, 4, 5, 8)$ is another solution).Listing all valid partitions of $20$ into $5$ distinct parts: $(1, 2, 3, 4, 10), (1, 2, 3, 5, 9), (1, 2, 3, 6, 8), (1, 2, 4, 5, 8), (1, 2, 4, 6, 7), (1, 3, 4, 5, 7), (2, 3, 4, 5, 6)$.There are exactly $7$ such arrangements.

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

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