Mathematics · Permutations and Combinations

JEE Main 2024 — 31 January, Shift 2 — Question 17

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{ }^{\mathrm{n}-1} \mathrm{P}_{3}:{ }^{\mathrm{n}} \mathrm{P}_{4}=1: 8, then nPm+1+n+1Cm{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}} is equal to

  1. Option A:

    380

  2. Option B:

    376

  3. Option C:

    384

  4. Option D:

    372

    Correct

Answer: D

Step-by-step solution

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

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

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

∴m=2\therefore \mathrm{m}=2

And 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\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}

∴n=8\therefore \mathrm{n}=8

∴nPm+1+n+1Cm=8P3+9C2\therefore{ }^{n} P_{m+1}+{ }^{n+1} C_{m}={ }^{8} P_{3}+{ }^{9} C_{2}

=8×7×6+9×82=8 \times 7 \times 6+\frac{9 \times 8}{2}

=372=372

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Combinations
If for some m, n ; 6 C m +2 ( 6 C m+1 )+ 6 C m+2 8 C 3 and n -1 P 3 … | JEE Main 2024 PYQ with Solution · DhiX AI