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Permutations and Combinations

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Question : 41 of 56

Marks: +1, -0

If

{ }^{n} c_{r-1}=56,{ }^{n} c_{r}=28

{ }^{n} c_{r+1}=8,

r

is equal to

8
6
5
none of these

Solution:

Concept:

Use ratio of consecutive combinations to relate

n

and

r

.

Explanation:

Given

{ }^{n}c_{r-1}=56

,

{ }^{n}c_{r}=28

,

{ }^{n}c_{r+1}=8

.

We know

\frac{{}^{n}c_{r}}{{}^{n}c_{r-1}} = \frac{n-r+1}{r}

.

So

\frac{28}{56} = \frac{n-r+1}{r} \implies \frac{1}{2} = \frac{n-r+1}{r}

.

Cross-multiply:

r = 2(n-r+1) \implies r = 2n-2r+2 \implies 3r = 2n+2

...(1)

Also

\frac{{}^{n}c_{r+1}}{{}^{n}c_{r}} = \frac{n-r}{r+1}

.

So

\frac{8}{28} = \frac{n-r}{r+1} \implies \frac{2}{7} = \frac{n-r}{r+1}

.

Cross-multiply:

2(r+1) = 7(n-r) \implies 2r+2 = 7n-7r \implies 9r = 7n-2

...(2)

From (1):

2n = 3r-2 \implies n = \frac{3r-2}{2}

.

Substitute into (2):

9r = 7\left(\frac{3r-2}{2}\right)-2

.

Multiply by 2:

18r = 7(3r-2)-4 \implies 18r = 21r-14-4 \implies 18r = 21r-18

.

Thus

3r = 18 \implies r = 6

.

Check:

n = \frac{3(6)-2}{2}=8

, and

{}^{8}c_{5}=56

,

{}^{8}c_{6}=28

,

{}^{8}c_{7}=8

confirm.

Answer:

r = 6

, which corresponds to option B.

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