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Exercise 6.4 · Q2

Q.Determine nn if

(i) 2nC3:nC3=12:1{}^{2n}C_3 : {}^{n}C_3 = 12 : 1
(ii) 2nC3:nC3=11:1{}^{2n}C_3 : {}^{n}C_3 = 11 : 1
Rajasthan RbseTextbookSubjective· 3mImportance★★★★★est
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✓ Free question

The key idea is to use the symmetry property of combinations and the given ratio to set up an equation in nn. For (i), the ratio 12:112:1 gives n=5n = 5; for (ii), the ratio 11:111:1 gives n=6n = 6.

We are given two separate problems involving the ratio of two combinations. The core concept here is the symmetry property of combinations: nCr=nCn−r{}^{n}C_{r} = {}^{n}C_{n-r}, but more importantly, we need the formula nCr=n!r!(n−r)!{}^{n}C_{r} = \frac{n!}{r!(n-r)!} to express both terms in the ratio. The ratio condition then becomes an algebraic equation in nn, which we solve.

Let’s work through each part step by step.

Part (i): 2nC3:nC3=12:1{}^{2n}C_3 : {}^{n}C_3 = 12 : 1

  1. Write the combinations in factorial form.

2nC3=(2n)!3!(2n−3)!,nC3=n!3!(n−3)!{}^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!}, \quad {}^{n}C_3 = \frac{n!}{3!(n-3)!}

  1. Set up the ratio equation. The given ratio is:

2nC3nC3=121\frac{{}^{2n}C_3}{{}^{n}C_3} = \frac{12}{1}

Substituting the expressions:

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

The 3!3! cancels out, leaving:

(2n)!(2n−3)!⋅(n−3)!n!=12\frac{(2n)!}{(2n-3)!} \cdot \frac{(n-3)!}{n!} = 12

  1. Simplify the factorials. Recall that (2n)!=(2n)(2n−1)(2n−2)(2n−3)!(2n)! = (2n)(2n-1)(2n-2)(2n-3)!, so:

(2n)!(2n−3)!=(2n)(2n−1)(2n−2)\frac{(2n)!}{(2n-3)!} = (2n)(2n-1)(2n-2)

Similarly, n!=n(n−1)(n−2)(n−3)!n! = n(n-1)(n-2)(n-3)!, so:

(n−3)!n!=1n(n−1)(n−2)\frac{(n-3)!}{n!} = \frac{1}{n(n-1)(n-2)}

Thus the equation becomes:

(2n)(2n−1)(2n−2)n(n−1)(n−2)=12\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 12

  1. Factor common terms. Notice 2n−2=2(n−1)2n-2 = 2(n-1). So the numerator is:

(2n)(2n−1)⋅2(n−1)=2n⋅2(n−1)⋅(2n−1)=4n(n−1)(2n−1)(2n)(2n-1) \cdot 2(n-1) = 2n \cdot 2(n-1) \cdot (2n-1) = 4n(n-1)(2n-1)

The denominator is n(n−1)(n−2)n(n-1)(n-2). Cancel n(n−1)n(n-1) (valid since n≥3n \ge 3 for combinations to be defined):

4(2n−1)n−2=12\frac{4(2n-1)}{n-2} = 12

  1. Solve for nn. Multiply both sides by n−2n-2:

4(2n−1)=12(n−2)4(2n-1) = 12(n-2)

Expand:

8n−4=12n−248n - 4 = 12n - 24

Bring terms together:

−4+24=12n−8n⇒20=4n-4 + 24 = 12n - 8n \quad \Rightarrow \quad 20 = 4n

So n=5n = 5.

Watch out

A common mistake is to forget that nn must be an integer and at least 3 (since nC3{}^{n}C_3 requires n≥3n \ge 3). Here n=5n=5 satisfies that.

Part (ii): 2nC3:nC3=11:1{}^{2n}C_3 : {}^{n}C_3 = 11 : 1

  1. Follow the same steps as above. The ratio equation is:

(2n)(2n−1)(2n−2)n(n−1)(n−2)=11\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 11

  1. Simplify identically. Cancelling n(n−1)n(n-1) as before gives:

4(2n−1)n−2=11\frac{4(2n-1)}{n-2} = 11

  1. Solve for nn. Multiply:

4(2n−1)=11(n−2)4(2n-1) = 11(n-2)

Expand:

8n−4=11n−228n - 4 = 11n - 22

Rearrange:

−4+22=11n−8n⇒18=3n-4 + 22 = 11n - 8n \quad \Rightarrow \quad 18 = 3n

So n=6n = 6.

Tip

Notice the pattern: the ratio 12:112:1 gave n=5n=5, and 11:111:1 gave n=6n=6. This is because the expression 4(2n−1)n−2\frac{4(2n-1)}{n-2} decreases as nn increases, so a smaller ratio corresponds to a larger nn.

✓Final answer

For (i), n=5n = 5; for (ii), n=6n = 6.

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