If ^2nC3^nC3 = 101, then the ratio (n^2 + 3n): (n^2 - 3n + 4) is: — Solution | JEE Mathematics

MCQEasyJEE 2023Combinations (C(n,r))

If $\frac{^{2n}C_{3}}{^{n}C_{3}} = \frac{10}{1}$ , then the ratio $(n^{2} + 3n) : (n^{2} - 3n + 4)$ is:

Answer

Correct answer:C

Standard Method

Given: $\frac{^{2n}C_{3}}{^{n}C_{3}} = 10$

Find: The ratio $(n^{2} + 3n) : (n^{2} - 3n + 4)$

Using combinations,

\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}=10

So,

\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}=10

Cancelling common factors,

\frac{4(2n-1)}{n-2}=10

Therefore,

4(2n-1)=10(n-2)

8n-4=10n-20

2n=16

n=8

Now substitute in the required ratio:

\frac{n^{2}+3n}{n^{2}-3n+4} = \frac{8^{2}+3(8)}{8^{2}-3(8)+4}

=\frac{64+24}{64-24+4}

=\frac{88}{44}=2

Hence, the ratio is $2:1$ . Therefore, the correct option is C.

Direct Simplification

Given: $\frac{^{2n}C_{3}}{^{n}C_{3}} = 10$

Find: The ratio $(n^{2} + 3n) : (n^{2} - 3n + 4)$

Write the combinations in product form:

\frac{^{2n}C_{3}}{^{n}C_{3}}=\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}

Since $2n-2=2(n-1)$ ,

\frac{2n(2n-1)\cdot 2(n-1)}{n(n-1)(n-2)}=10

Cancelling $n$ and $(n-1)$ ,

\frac{4(2n-1)}{n-2}=10

Now solve:

4(2n-1)=10(n-2)

8n-4=10n-20

16=2n

n=8

Then,

n^{2}+3n=64+24=88

and

n^{2}-3n+4=64-24+4=44

So the ratio is

88:44=2:1

Thus, the required ratio is $2:1$ .

Cancelling terms incorrectly in $\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}$ is a common mistake. $2n-2$ should be written as $2(n-1)$ before cancellation. Do the factorization first, then cancel only common factors.
Treating the final value $\frac{88}{44}=2$ as the ratio itself without converting it properly can confuse students. Since the expression asks for a ratio, the correct form is $2:1$ , not only $2$ .
Using the wrong formula for combinations, such as expanding factorials incompletely, leads to algebra errors. Write $^{n}C_{3}=\frac{n(n-1)(n-2)}{3!}$ carefully before simplifying.

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