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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2024 · 31 January · Evening Shift · Mathematics · Binomial Theorem

If for some m, n; ^ 6 C _ m + 2( ^ 6 C _ m + 1 ) + ^ 6 C _ m + 2 >...

If for some m,n;6Cm+2(6Cm+1)+6Cm+2>8C3m, n; {}^{6}{C}_{m} + 2\left({}^{6}{C}_{m + 1}\right){+}^{6}{C}_{m + 2}{>}^{8}{C}_{3} and n−1P3:nP4=1:8{}^{n - 1}{P}_{3}{:}^{n}{P}_{4} = 1: 8, then nPm+1+n+1Cm{}^{n}{P}_{m + 1}{+}^{n + 1}{C}_{m} is equal to

Options

  1. A380380
  2. B376376
  3. C384384
  4. D372372

Correct answer

D. 372372

Step-by-step solution

Given: 6Cm+2(6Cm+1)+6Cm+2>8C3{}^{6}{C}_{m} + 2\left({}^{6}{C}_{m + 1}\right){+}^{6}{C}_{m + 2}{>}^{8}{C}_{3}

⇒6Cm+6Cm+1+6Cm+1+6Cm+2>8C3{⇒}^{6}{C}_{m}{+}^{6}{C}_{m + 1}{+}^{6}{C}_{m + 1}{+}^{6}{C}_{m + 2}{>}^{8}{C}_{3}

⇒7Cm+1+7Cm+2>8C3{⇒}^{7}{C}_{m + 1}{+}^{7}{C}_{m + 2}{>}^{8}{C}_{3}

⇒8Cm+2>8C3{⇒}^{8}{C}_{m + 2}{>}^{8}{C}_{3}

⇒m=2⇒ m = 2

Also, n−1P3:nP4=1:8{}^{n - 1}{P}_{3}{:}^{n}{P}_{4} = 1: 8

⇒(n−1)(n−2)(n−3)n(n−1)(n−2)(n−3)=18⇒ \dfrac{\left(n - 1\right)\left(n - 2\right)\left(n - 3\right)}{n\left(n - 1\right)\left(n - 2\right)\left(n - 3\right)} = \dfrac{1}{8}

⇒n=8⇒ n = 8

⇒nPm+1+n+1Cm=8P3+9C2{⇒}^{n}{P}_{m + 1}{+}^{n + 1}{C}_{m}{=}^{8}{P}_{3}{+}^{9}{C}_{2}

⇒nPm+1+n+1Cm=8×7×6+9×82{⇒}^{n}{P}_{m + 1}{+}^{n + 1}{C}_{m} = 8 \times 7 \times 6 + \dfrac{9 \times 8}{2}

⇒nPm+1+n+1Cm=372{⇒}^{n}{P}_{m + 1}{+}^{n + 1}{C}_{m} = 372

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