Let $\binom{n}{k}$ denote ${ }^{n} C_{k}$ and $\left[\begin{array}{c}n \\ k\end{array}\right]=\left\{\begin{array}{l}\binom{n}{k} \text {, if } 0 \leq k \leq n \\ 0 \text {, otherwise }\end{array}\right.$ If $\mathrm{A}_{\mathrm{k}}=\sum_{\mathrm{i}=0}^{9}\binom{9}{\mathrm{i}}\left[\begin{array}{c}12 \\ 12-\mathrm{k}+\mathrm{i}\end{array}\right]+\sum_{\mathrm{i}=0}^{8}\binom{8}{\mathrm{i}}\left[\begin{array}{c}13 \\ 13-\mathrm{k}+\mathrm{i}\end{array}\right]$ and $\mathrm{A}_{4}-\mathrm{A}_{3}=190 \mathrm{p}$, then p is equal to ______.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Binomial Theorem

Let

(k n)

denote

^{n} C_{k}

and

[n k] = {(k n), if 0 \leq k \leq n 0, otherwise

A_{k} = \sum_{i = 0}^{9} (i 9) [12 12 - k + i] + \sum_{i = 0}^{8} (i 8) [13 13 - k + i]

and

A_{4} - A_{3} = 190 p

, then p is equal to ______.

Answer: 49

Solution

In the first sum replace

(i 9)

(9 - i 9)

and put

j = 9 - i

i = 0 \sum 9 (i 9) [12 12 - k + i] = j = 0 \sum 9 (j 9) (( 21 - k ) - j 12) = (21 - k 21) = (k 21)

by Vandermonde's identity, the bracket symbol being exactly the device that makes the out-of-range terms vanish. Treating the second sum the same way with

j = 8 - i

i = 0 \sum 8 (i 8) [13 13 - k + i] = j = 0 \sum 8 (j 8) (( 21 - k ) - j 13) = (21 - k 21) = (k 21)

Hence

A_{k} = 2 (k 21)

, and

A_{4} - A_{3} = 2 [(4 21) - (3 21)] = 2 (5985 - 1330) = 9310

Since

9310 = 190 \times 49

, the value of

p

49

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Binomial Theorem

Solution

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