If for some \(\mathrm{m}, \mathrm{n} ;{ }^{6} \mathrm{C}_{\mathrm{m}}+2\left({ }^{6} \mathrm{C}_{\mathrm{m}+1}\right)+{ }^{6} \mathrm{C}_{\mathrm{m}+2}>{ }^{8} \mathrm{C}_{3}\) and \({ }^{n-1} P_{3}:{ }^{n} P_{4}=1: 8\), then \({ }^{n} P_{m+1}+{ }^{n+1} C_{m}\) is equal to

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Permutations and 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} :^{n} P_{4} = 1 : 8

, then

^{n} P_{m + 1} +^{n + 1} C_{m}

is equal to

Answer: (D)

^{6} C_{m} + 2^{6} C_{m + 1} +^{6} C_{m + 2} = (^{6} C_{m} +^{6} C_{m + 1}) + (^{6} C_{m + 1} +^{6} C_{m + 2}) =^{7} C_{m + 1} +^{7} C_{m + 2} =^{8} C_{m + 2}

(using Pascal's rule twice). The condition becomes

^{8} C_{m + 2} >^{8} C_{3}

.

Since

^{8} C_{r}

increases up to

r = 4

and

^{8} C_{3} =^{8} C_{5} = 56 <^{8} C_{4} = 70

, the only value of

m + 2

that makes

^{8} C_{m + 2}

strictly greater than

^{8} C_{3}

m + 2 = 4

, so

m = 2

.

For

n

^{n - 1} P_{3} :^{n} P_{4} = 1 : 8

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

.

So

^{n} P_{m + 1} +^{n + 1} C_{m} =^{8} P_{3} +^{9} C_{2} = 8 \times 7 \times 6 + \frac{9 \times 8}{2} = 336 + 36 = 372

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option