If 2nC₃: nC₂=44 :3, then the value of n is

None of the above/More than one of the above

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BPSC 66th Prelims 27 Dec 2020 Paper

Question : 142 of 150

Marks: +1, -0

2nC_{3}: nC_{2}=44 :3

, then the value of n is

Solution:

Concept:

Use the combination formula

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

to write the given ratio as an equation and solve for

n

Explanation:

Write

^{2n}C_3

and

^nC_2

in expanded form.

^{2n}C_3 = \frac{(2n)(2n-1)(2n-2)}{3!}

^nC_2 = \frac{n(n-1)}{2!}

The given ratio is

^{2n}C_3 : ^nC_2 = 44 : 3

So,

\frac{^{2n}C_3}{^nC_2} = \frac{44}{3}

Substitute the expressions:

\frac{(2n)(2n-1)(2n-2)/3!}{n(n-1)/2!} = \frac{44}{3}

Simplify:

\frac{(2n)(2n-1)(2n-2) \cdot 2!}{n(n-1) \cdot 3!} = \frac{44}{3}

Cancel

n

and simplify:

2n-2 = 2(n-1)

, so

\frac{2 \cdot (2n-1) \cdot 2}{3 \cdot 2?}

Let's do carefully:

\frac{(2n)(2n-1)(2n-2) \cdot 2}{n(n-1) \cdot 6} = \frac{44}{3}

Since

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

, cancel

n(n-1)

\frac{4 \cdot (2n-1) \cdot 2}{6} = \frac{44}{3}

→

\frac{8(2n-1)}{6} = \frac{44}{3}

→

\frac{4(2n-1)}{3} = \frac{44}{3}

Multiply both sides by 3:

4(2n-1) = 44

→

2n-1 = 11

→

2n = 12

→

n = 6

Answer:

n = 6

, which matches option A.

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