The key idea is to use the symmetry property of combinations and the given ratio to set up an equation in n. For (i), the ratio 12:1 gives n=5; for (ii), the ratio 11:1 gives n=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, but more importantly, we need the formula nCr=r!(n−r)!n! to express both terms in the ratio. The ratio condition then becomes an algebraic equation in n, which we solve.
Let’s work through each part step by step.
Part (i): 2nC3:nC3=12:1
- Write the combinations in factorial form.
2nC3=3!(2n−3)!(2n)!,nC3=3!(n−3)!n!
- Set up the ratio equation.
The given ratio is:
nC32nC3=112
Substituting the expressions:
3!(n−3)!n!3!(2n−3)!(2n)!=12
The 3! cancels out, leaving:
(2n−3)!(2n)!⋅n!(n−3)!=12
- Simplify the factorials.
Recall that (2n)!=(2n)(2n−1)(2n−2)(2n−3)!, so:
(2n−3)!(2n)!=(2n)(2n−1)(2n−2)
Similarly, n!=n(n−1)(n−2)(n−3)!, so:
n!(n−3)!=n(n−1)(n−2)1
Thus the equation becomes:
n(n−1)(n−2)(2n)(2n−1)(2n−2)=12
- Factor common terms.
Notice 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)
The denominator is n(n−1)(n−2). Cancel n(n−1) (valid since n≥3 for combinations to be defined):
n−24(2n−1)=12
- Solve for n.
Multiply both sides by n−2:
4(2n−1)=12(n−2)
Expand:
8n−4=12n−24
Bring terms together:
−4+24=12n−8n⇒20=4n
So n=5.
A common mistake is to forget that n must be an integer and at least 3 (since nC3 requires n≥3). Here n=5 satisfies that.
Part (ii): 2nC3:nC3=11:1
- Follow the same steps as above.
The ratio equation is:
n(n−1)(n−2)(2n)(2n−1)(2n−2)=11
- Simplify identically.
Cancelling n(n−1) as before gives:
n−24(2n−1)=11
- Solve for n.
Multiply:
4(2n−1)=11(n−2)
Expand:
8n−4=11n−22
Rearrange:
−4+22=11n−8n⇒18=3n
So n=6.
Notice the pattern: the ratio 12:1 gave n=5, and 11:1 gave n=6. This is because the expression n−24(2n−1) decreases as n increases, so a smaller ratio corresponds to a larger n.
✓Final answer
For (i), n=5; for (ii), n=6.