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Permutations and Combinations

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Question : 28 of 56

Marks: +1, -0

If

{ }^{n} P_{13}:{ }^{n+1} P_{12}=3: 4

n

is

15
14
13
12

Solution:

Concept:

Use the permutation formula

^nP_r = \frac{n!}{(n-r)!}

and simplify the ratio to solve for

n

.

Explanation:

Given

\frac{^nP_{13}}{^{n+1}P_{12}} = \frac{3}{4}

.

Write

^nP_{13} = \frac{n!}{(n-13)!}

and

^{n+1}P_{12} = \frac{(n+1)!}{(n-11)!}

.

Then

\frac{n!}{(n-13)!} \cdot \frac{(n-11)!}{(n+1)!} = \frac{3}{4}

.

Since

(n+1)! = (n+1) \cdot n!

and

(n-11)! = (n-11)(n-12)(n-13)!

, the left side simplifies to

\frac{(n-11)(n-12)}{n+1}

.

Thus

\frac{(n-11)(n-12)}{n+1} = \frac{3}{4}

.

Cross-multiply:

4(n-11)(n-12) = 3(n+1)

.

Expand:

4(n^2 - 23n + 132) = 3n + 3 \implies 4n^2 - 92n + 528 = 3n + 3

.

Simplify:

4n^2 - 95n + 525 = 0

.

Solve:

n = \frac{95 \pm \sqrt{9025 - 8400}}{8} = \frac{95 \pm 25}{8}

.

So

n = \frac{120}{8} = 15

or

n = \frac{70}{8} = 8.75

(invalid).

Thus

n = 15

.

Answer:

n = 15

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