If and are in the ratio , then the value of is:

5 Explanation: 1. Understand the Notation The expressions given are the formulas for combinations: The problem states that . 2. Set up the Equation 3. Simplify the Fraction Flip and multiply the denominator: 4. Expand the Factorials Expand and to cancel out terms: 5. Solve for Divide both sides by : Expand the quadratic: 6. Conclusion The possible values for are or . However, for the combination to be defined, must be at least . Therefore, . Correct Option: (D)

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NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$ , then the value of $n$ is:

Choose the correct answer:

Explanation

1. Understand the Notation

The expressions given are the formulas for combinations:

$\frac{n!}{2!(n-2)!} = {}^nC_2$

$\frac{n!}{4!(n-4)!} = {}^nC_4$

The problem states that ${}^nC_2 : {}^nC_4 = 2 : 1$ .

2. Set up the Equation

$\frac{{}^nC_2}{{}^nC_4} = \frac{2}{1}$

$\frac{\frac{n!}{2!(n-2)!}}{\frac{n!}{4!(n-4)!}} = 2$

3. Simplify the Fraction

Flip and multiply the denominator:

$\frac{n!}{2!(n-2)!} \times \frac{4!(n-4)!}{n!} = 2$

$\frac{4!(n-4)!}{2!(n-2)!} = 2$

4. Expand the Factorials

Expand $4!$ and $(n-2)!$ to cancel out terms:

$\frac{(4 \times 3 \times 2!) \times (n-4)!}{2! \times (n-2)(n-3)(n-4)!} = 2$

$\frac{4 \times 3}{(n-2)(n-3)} = 2$

5. Solve for $n$

$\frac{12}{(n-2)(n-3)} = 2$

Divide both sides by $2$ :

$\frac{6}{(n-2)(n-3)} = 1$

$(n-2)(n-3) = 6$

Expand the quadratic:

$n^2 - 5n + 6 = 6$

$n^2 - 5n = 0$

$n(n - 5) = 0$

6. Conclusion

The possible values for $n$ are $0$ or $5$ . However, for the combination ${}^nC_4$ to be defined, $n$ must be at least $4$ . Therefore, $n = 5$ .

Correct Option: (D)