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QJEEM24Ma9fe5: JEE Main 2024 Mathematics - Class 11 JEEM24Ma9fe5 Solution: If for some and , then is equal to

If for some m, 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}+{ }^{\mathrm{n}+1} C_m is equal to

Multiple Choice Options

  • A 380
  • B 376
  • C 372
  • D 384

Step-by-Step Verified Solution

amp;6Cm+2(6Cm+1)+6Cm+2gt;8C3amp;7Cm+1+7Cm+2gt;8C3amp;8Cm+2gt;8C3amp;∴m=2amp; And n−1P3:nP4=1:8amp;(n−1)(n−2)(n−3)n(n−1)(n−2)(n−3)=18amp;∴n=8amp;∴nPm+1+n+1Cm=8P3+9C2amp;=8×7×6+9×82amp;=372\begin{aligned}& { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\& { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\& { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\& \therefore \mathrm{m}=2 \\& \text { And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 8 \\& \frac{(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=\frac{1}{8} \\& \therefore \mathrm{n}=8 \\& \therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2 \\& =8 \times 7 \times 6+\frac{9 \times 8}{2} \\& =372\end{aligned}

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