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Question: If \[^{n}{{C}_{r-1}}=36,{{\text{ }}^{n}}{{C}_{r}}=84\text{ and}{{\text{ }}^{n}}{{C}_{r+1}}=126\], th...

If nCr−1=36, nCr=84 and nCr+1=126^{n}{{C}_{r-1}}=36,{{\text{ }}^{n}}{{C}_{r}}=84\text{ and}{{\text{ }}^{n}}{{C}_{r+1}}=126, then the value of nC8^{n}{{C}_{8}} is:
(a) 10
(b) 7
(c) 9
(d) 8

Explanation

Solution

Hint: Here use the formula of nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}. Replace r with (r – 1) and (r + 1) to find the value of nCr−1^{n}{{C}_{r-1}} and nCr+1^{n}{{C}_{r+1}}. Divide these values to get the value of n. Then find the value of nC8^{n}{{C}_{8}}.

Complete step-by-step answer:

Here, we are given that nCr−1=36, nCr=84 and nCr+1=126^{n}{{C}_{r-1}}=36,{{\text{ }}^{n}}{{C}_{r}}=84\text{ and}{{\text{ }}^{n}}{{C}_{r+1}}=126. We have to find the value of nC8^{n}{{C}_{8}}.
First of all, let us see the formula for nCr^{n}{{C}_{r}}.
We have nCr=n!r!(n−r)!....(i)^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}....\left( i \right)
We are given that nCr=84^{n}{{C}_{r}}=84. By substituting the value of nCr^{n}{{C}_{r}} in the above equation, we get,
n!r!(n−r)!=84....(ii)\dfrac{n!}{r!\left( n-r \right)!}=84....\left( ii \right)
Now by replacing (r) by (r – 1) in equation (i), we get,
nC(r−1)=n!(r−1)!(n−(r−1))!^{n}{{C}_{\left( r-1 \right)}}=\dfrac{n!}{\left( r-1 \right)!\left( n-\left( r-1 \right) \right)!}
Or, nC(r−1)=n!(r−1)!(n−r+1)!^{n}{{C}_{\left( r-1 \right)}}=\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}
We are given that nC(r−1)=36^{n}{{C}_{\left( r-1 \right)}}=36. By substituting the value of nC(r−1)^{n}{{C}_{\left( r-1 \right)}} in the above equation, we get,
n!(r−1)!(n−r+1)!=36....(iii)\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}=36....\left( iii \right)
Now, by replacing (r) by (r + 1) in equation (i), we get,
nC(r+1)=n!(r+1)!(n−(r+1))!^{n}{{C}_{\left( r+1 \right)}}=\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}
Or, nC(r+1)=n!(r+1)!(n−r−1)!^{n}{{C}_{\left( r+1 \right)}}=\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}
We are given that nC(r+1)=126^{n}{{C}_{\left( r+1 \right)}}=126. By substituting the value of nC(r+1)^{n}{{C}_{\left( r+1 \right)}} in the above equation, we get,
n!(r+1)!(n−r−1)!=126....(iv)\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}=126....\left( iv \right)
Now, by dividing equation (ii) and (iii), we get,
n!r!(n−r)!n!(r−1)!(n−r+1)!=8436\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}}=\dfrac{84}{36}
By canceling the like terms and simplifying the above equation, we get,
(r−1)!(n−r+1)!r!(n−r)!=73\dfrac{\left( r-1 \right)!\left( n-r+1 \right)!}{r!\left( n-r \right)!}=\dfrac{7}{3}
We know that m!=(m).(m−1).(m−2).(m−3).(m−4).....3.2.1m!=\left( m \right).\left( m-1 \right).\left( m-2 \right).\left( m-3 \right).\left( m-4 \right).....3.2.1.
So, here we can write r!=r.(r−1)!r!=r.\left( r-1 \right)! and (n−r+1)!=(n−r+1).(n−r+1−1)!=(n−r+1)(n−r)!\left( n-r+1 \right)!=\left( n-r+1 \right).\left( n-r+1-1 \right)!=\left( n-r+1 \right)\left( n-r \right)!
By substituting the values of r! and (n – r + 1)! in the above equation, we get,
(r−1)!(n−r+1).(n−r)!(r).(r−1)!.(n−r)!=73\dfrac{\left( r-1 \right)!\left( n-r+1 \right).\left( n-r \right)!}{\left( r \right).\left( r-1 \right)!.\left( n-r \right)!}=\dfrac{7}{3}
By canceling the like terms, we get,
n−r+1r=73\dfrac{n-r+1}{r}=\dfrac{7}{3}
By cross multiplying the above equation, we get,
3n−3r+3=7r3n-3r+3=7r
Or, 7r+3r−3n=37r+3r-3n=3
⇒10r−3n=3....(v)\Rightarrow 10r-3n=3....\left( v \right)
Now, by dividing equation (ii) and (iv), we get,
n!r!(n−r)!n!(r+1)!(n−r−1)!=84126\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}}=\dfrac{84}{126}
By canceling the like terms and simplifying the above equation, we get,
(r+1)!(n−r−1)!r!(n−r)!=23\dfrac{\left( r+1 \right)!\left( n-r-1 \right)!}{r!\left( n-r \right)!}=\dfrac{2}{3}
Here, we can write (r+1)!=(r+1).(r+1−1)!=(r+1)(r)!\left( r+1 \right)!=\left( r+1 \right).\left( r+1-1 \right)!=\left( r+1 \right)\left( r \right)! and (n−r)!=(n−r)(n−r−1)!\left( n-r \right)!=\left( n-r \right)\left( n-r-1 \right)!.
By substituting the values of (r + 1)! and (n – r)! in the above equation, we get,
(r+1)(r)!(n−r−1)!(r)!(n−r)(n−r−1)!=23\dfrac{\left( r+1 \right)\left( r \right)!\left( n-r-1 \right)!}{\left( r \right)!\left( n-r \right)\left( n-r-1 \right)!}=\dfrac{2}{3}
By canceling the like terms, we get,
r+1n−r=23\dfrac{r+1}{n-r}=\dfrac{2}{3}
By cross multiplying the above equation, we get,
3(r+1)=2(n−r)3\left( r+1 \right)=2\left( n-r \right)
Or, 3r+3=2n−2r3r+3=2n-2r
Or, 2n−2r−3r=32n-2r-3r=3
⇒2n−5r=3\Rightarrow 2n-5r=3
By multiplying the above equation by 2 on both sides, we get,
4n−10r=6....(vi)4n-10r=6....\left( vi \right)
Now by adding equations (v) and (vi), we get,
⇒(10r−3n)+(4n−10r)=3+6\Rightarrow \left( 10r-3n \right)+\left( 4n-10r \right)=3+6
Or, −3n+4n=9-3n+4n=9
Therefore we get n = 9.
Now, to find nC8^{n}{{C}_{8}}, we will substitute r = 8 in equation (i), we get,
nC8=n!8!(n−8)!^{n}{{C}_{8}}=\dfrac{n!}{8!\left( n-8 \right)!}
By substituting the value of n = 9 in the above equation, we get,
nC8= 9C8=9!8!(9−8)!^{n}{{C}_{8}}={{\text{ }}^{9}}{{C}_{8}}=\dfrac{9!}{8!\left( 9-8 \right)!}
Here, we can write 9! = 9 x 8!. So, we get,
nC8= 9C8=9×8!8!1!^{n}{{C}_{8}}={{\text{ }}^{9}}{{C}_{8}}=\dfrac{9\times 8!}{8!1!}
By canceling the like terms, we get,
nC8= 9C8=9^{n}{{C}_{8}}={{\text{ }}^{9}}{{C}_{8}}=9
Hence, the option (c) is the right answer.

Note: In this question, after getting the value of n, many students substitute it back in the equation to get the value of r. But they must note that the value of r is not required here. So they should not waste their time unnecessarily in finding the value of r. Also, students are required to remember the formula for nCr^{n}{{C}_{r}} that is n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!} as it is a very useful formula for mathematics.