JEE Mathematics Question

Given: ^n-1Cr = (k^2 - 8) ^nCr+1 Find: the range of k. Using the combination formulas, ^n-1Cr = (n-1)!r!(n-r-1)!, ^nCr+1 = n!(r+1)!(n-r-1)! Substitute these in the given relation: (n-1)!r!(n-r-1)! = (k^2 - 8) n!(r+1)!(n-r-1)! Cancelling common factors gives k^2 - 8 = r+1n Since for valid combinations we have 0 r n-1, therefore 0 < r+1 n Hence, 0 < r+1n 1 So, 0 < k^2 - 8 1 This gives 8 < k^2 9 Therefore, taking the positive range indicated by the options, 2 2 < k 3 So the correct option is A. The solution contains an intermediate simplification to 1n, but the consistent factorial simplification yields r+1n, which still leads to the same listed option A. Given: ^n-1Cr = (k^2 - 8) ^nCr+1 Find: the admissible values of k. Rewrite the factor multiplying ^nCr+1 as a ratio: k^2 - 8 = ^n-1Cr^nCr+1 Now evaluate this ratio directly: ^n-1Cr^nCr+1 = (n-1)!r!(n-r-1)! n!(r+1)!(n-r-1)! = (n-1)!(r+1)!r! n! = r+1n Because r is an integer with 0 r n-1, 1 r+1 n Dividing by n gives 1n r+1n 1 In particular, it is positive and at most 1. Hence, 0 < k^2 - 8 1 8 < k^2 9 Thus, 2 2 < k 3 Among the given choices, this is option A.

JEE Question — Solution | JEE Mathematics

MCQMediumJEE 2024Combinations (C(n,r))

$Equation image showing $$^{n-1}C_r = (k^2 - 8)\, ^nC_{r+1}$$ followed by the words if and only if:$

A
$2\sqrt{2} < k \le 3$
B
$2\sqrt{3} < k \le 3\sqrt{2}$
C
$2\sqrt{3} < k < 3\sqrt{3}$
D
$2\sqrt{2} < k < 2\sqrt{3}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

{}^{n-1}C_r = (k^2 - 8)\, {}^nC_{r+1}

Find: the range of $k$ .

Using the combination formulas,

{}^{n-1}C_r = \frac{(n-1)!}{r!(n-r-1)!}, \qquad {}^nC_{r+1} = \frac{n!}{(r+1)!(n-r-1)!}

Substitute these in the given relation:

\frac{(n-1)!}{r!(n-r-1)!} = (k^2 - 8) \cdot \frac{n!}{(r+1)!(n-r-1)!}

Cancelling common factors gives

k^2 - 8 = \frac{r+1}{n}

Since for valid combinations we have $0 \le r \le n-1$ , therefore

0 < r+1 \le n

Hence,

0 < \frac{r+1}{n} \le 1

So,

0 < k^2 - 8 \le 1

This gives

8 < k^2 \le 9

Therefore, taking the positive range indicated by the options,

2\sqrt{2} < k \le 3

So the correct option is A.

The solution contains an intermediate simplification to $\frac{1}{n}$ , but the consistent factorial simplification yields $\frac{r+1}{n}$ , which still leads to the same listed option A.

Range-Based Interpretation

Given:

{}^{n-1}C_r = (k^2 - 8)\, {}^nC_{r+1}

Find: the admissible values of $k$ .

Rewrite the factor multiplying ${}^nC_{r+1}$ as a ratio:

k^2 - 8 = \frac{{}^{n-1}C_r}{{}^nC_{r+1}}

Now evaluate this ratio directly:

\frac{{}^{n-1}C_r}{{}^nC_{r+1}} = \frac{\frac{(n-1)!}{r!(n-r-1)!}}{\frac{n!}{(r+1)!(n-r-1)!}} = \frac{(n-1)!(r+1)!}{r!\, n!} = \frac{r+1}{n}

Because $r$ is an integer with $0 \le r \le n-1$ ,

1 \le r+1 \le n

Dividing by $n$ gives

\frac{1}{n} \le \frac{r+1}{n} \le 1

In particular, it is positive and at most $1$ .

Hence,

0 < k^2 - 8 \le 1 \Rightarrow 8 < k^2 \le 9

Thus,

2\sqrt{2} < k \le 3

Among the given choices, this is option A.

Common mistakes

A common mistake is simplifying the factorial ratio to $\frac{1}{n}$ instead of $\frac{r+1}{n}$ . This is wrong because $(r+1)!/r! = r+1$ . Always cancel factorials carefully before concluding the range.
Another mistake is ignoring the condition on $r$ in combinations. Since $0 \le r \le n-1$ , we must use $1 \le r+1 \le n$ . Without this, the upper bound on $k$ cannot be obtained correctly.
Students may take $k^2 > 8$ and stop at $k > 2\sqrt{2}$ . This misses the additional fact that $k^2 - 8 = \frac{r+1}{n} \le 1$ , which gives the upper bound $k \le 3$ .

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Step-by-step solution

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