JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 16)
If for some $$m, n ;{ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3$$ and $${ }^{n-1} P_3:{ }^n P_4=1: 8$$, then $${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$$ is equal to
380
376
372
384
Explanation
$$\begin{aligned}
& { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& \therefore \mathrm{m}=2 \\
& \text { And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 8 \\
& \frac{(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=\frac{1}{8} \\
& \therefore \mathrm{n}=8 \\
& \therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2 \\
& =8 \times 7 \times 6+\frac{9 \times 8}{2} \\
& =372
\end{aligned}$$
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