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Question: If \[{}^{2n}{c_3}:{}^n{c_2} = 44:3\] then for which of the following value of \[r\] , the value of \...

If 2nc3:nc2=44:3{}^{2n}{c_3}:{}^n{c_2} = 44:3 then for which of the following value of rr , the value of ncr{}^n{c_r} will be 1515?
A. r=3r = 3
B. r=4r = 4
C. r=6r = 6
D. r=5r = 5

Explanation

Solution

In the given problem, we have to find the value of rr. We will proceed with 2nc3:nc2=44:3{}^{2n}{c_3}:{}^n{c_2} = 44:3 and simplify to find the value of nn. We will use the formula ncr=n!r!(n−r)!{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} in above and solve it to get the value of nn. We will then compare ncr{}^n{c_r} to 6c2=15{}^6{c_2} = 15 to get the value of rr.

Complete step by step answer:
Consider the given question,
We are given , 2nc3:nc2=44:3{}^{2n}{c_3}:{}^n{c_2} = 44:3.
i.e. 2nc3nc2=443\dfrac{{{}^{2n}{c_3}}}{{{}^n{c_2}}} = \dfrac{{44}}{3}
Cross multiplying and then putting the value of 2nc3{}^{2n}{c_3} and nc2{}^n{c_2} we have,
⇒3×2n!3!(2n−3)!=44×n!2!(n−2)!\Rightarrow 3 \times \dfrac{{2n!}}{{3!\left( {2n - 3} \right)!}} = 44 \times \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}
On simplifying, we have
⇒3×2n(2n−1)(2n−2)(2n−3)!3!(2n−3)!=44×n(n−1)(n−2)!2!(n−2)!\Rightarrow 3 \times \dfrac{{2n(2n - 1)(2n - 2)(2n - 3)!}}{{3!\left( {2n - 3} \right)!}} = 44 \times \dfrac{{n(n - 1)(n - 2)!}}{{2!\left( {n - 2} \right)!}}
Cancelling (2n−3)!(2n - 3)! in LHS and (n−2)!(n - 2)! in RHS we have,
⇒3×2n(2n−1)(2n−2)3!=44×n(n−1)2!\Rightarrow 3 \times \dfrac{{2n(2n - 1)(2n - 2)}}{{3!}} = 44 \times \dfrac{{n(n - 1)}}{{2!}}
Taking common in LHS and on simplifying we have,
⇒3×2×2n(2n−1)(n−1)3×2×1=44×n(n−1)2×1\Rightarrow 3 \times \dfrac{{2 \times 2n(2n - 1)(n - 1)}}{{3 \times 2 \times 1}} = 44 \times \dfrac{{n(n - 1)}}{{2 \times 1}}

On simplifying, by cancelling n(n−1)n(n - 1) both side and the equal terms we get
⇒2(2n−1)=22\Rightarrow 2(2n - 1) = 22
On simplifying,
⇒4n−2=22\Rightarrow 4n - 2 = 22
Adding 22 both side, we get
⇒4n=24\Rightarrow 4n = 24
Dividing by 44 both side we get
⇒n=6\Rightarrow n = 6.
Now we have to find the value of rr, when ncr=15{}^n{c_r} = 15
Consider, ncr=15{}^n{c_r} = 15
We know that 6c2=15{}^6{c_2} = 15
Hence, 6cr=6c2{}^6{c_r} = {}^6{c_2}
Comparing, we have r=2r = 2 or 44.

Hence option B is correct.

Note: The factorial of any number is written as n!=n(n−1)(n−2)...........3×2×1n! = n(n - 1)(n - 2)...........3 \times 2 \times 1. We can also write n!=n×(n−1)!n! = n \times (n - 1)!, this will help to cancel the terms easily. The process of selecting r things out of nn thing is given by ncr=n!r!(n−r)!{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}. The process of arranging r things out of n things is given by npr=n!(n−r)!{}^n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}