Question

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

Answer 1

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

Complete step-by-step answer:

Here, we are given that \[^{n}{{C}_{r-1}}=36,{{\text{ }}^{n}}{{C}_{r}}=84\text{ and}{{\text{ }}^{n}}{{C}_{r+1}}=126\]. We have to find the value of \[^{n}{{C}_{8}}\].
First of all, let us see the formula for \[^{n}{{C}_{r}}\].
We have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}....\left( i \right)\]
We are given that \[^{n}{{C}_{r}}=84\]. By substituting the value of \[^{n}{{C}_{r}}\] in the above equation, we get,
\[\dfrac{n!}{r!\left( n-r \right)!}=84....\left( ii \right)\]
Now by replacing (r) by (r – 1) in equation (i), we get,
\[^{n}{{C}_{\left( r-1 \right)}}=\dfrac{n!}{\left( r-1 \right)!\left( n-\left( r-1 \right) \right)!}\]
Or, \[^{n}{{C}_{\left( r-1 \right)}}=\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}\]
We are given that \[^{n}{{C}_{\left( r-1 \right)}}=36\]. By substituting the value of \[^{n}{{C}_{\left( r-1 \right)}}\] in the above equation, we get,
\[\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}=36....\left( iii \right)\]
Now, by replacing (r) by (r + 1) in equation (i), we get,
\[^{n}{{C}_{\left( r+1 \right)}}=\dfrac{n!}{\left( r+1 \right)!\left( n-\left( r+1 \right) \right)!}\]
Or, \[^{n}{{C}_{\left( r+1 \right)}}=\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}\]
We are given that \[^{n}{{C}_{\left( r+1 \right)}}=126\]. By substituting the value of \[^{n}{{C}_{\left( r+1 \right)}}\] in the above equation, we get,
\[\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}=126....\left( iv \right)\]
Now, by dividing equation (ii) and (iii), we get,
\[\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r-1 \right)!\left( n-r+1 \right)!}}=\dfrac{84}{36}\]
By canceling the like terms and simplifying the above equation, we get,
\[\dfrac{\left( r-1 \right)!\left( n-r+1 \right)!}{r!\left( n-r \right)!}=\dfrac{7}{3}\]
We know that \[m!=\left( m \right).\left( m-1 \right).\left( m-2 \right).\left( m-3 \right).\left( m-4 \right).....3.2.1\].
So, here we can write \[r!=r.\left( r-1 \right)!\] and \[\left( n-r+1 \right)!=\left( n-r+1 \right).\left( n-r+1-1 \right)!=\left( n-r+1 \right)\left( n-r \right)!\]
By substituting the values of r! and (n – r + 1)! in the above equation, we get,
\[\dfrac{\left( r-1 \right)!\left( n-r+1 \right).\left( n-r \right)!}{\left( r \right).\left( r-1 \right)!.\left( n-r \right)!}=\dfrac{7}{3}\]
By canceling the like terms, we get,
\[\dfrac{n-r+1}{r}=\dfrac{7}{3}\]
By cross multiplying the above equation, we get,
\[3n-3r+3=7r\]
Or, \[7r+3r-3n=3\]
\[\Rightarrow 10r-3n=3....\left( v \right)\]
Now, by dividing equation (ii) and (iv), we get,
\[\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{\left( r+1 \right)!\left( n-r-1 \right)!}}=\dfrac{84}{126}\]
By canceling the like terms and simplifying the above equation, we get,
\[\dfrac{\left( r+1 \right)!\left( n-r-1 \right)!}{r!\left( n-r \right)!}=\dfrac{2}{3}\]
Here, we can write \[\left( r+1 \right)!=\left( r+1 \right).\left( r+1-1 \right)!=\left( r+1 \right)\left( r \right)!\] and \[\left( n-r \right)!=\left( n-r \right)\left( n-r-1 \right)!\].
By substituting the values of (r + 1)! and (n – r)! in the above equation, we get,
\[\dfrac{\left( r+1 \right)\left( r \right)!\left( n-r-1 \right)!}{\left( r \right)!\left( n-r \right)\left( n-r-1 \right)!}=\dfrac{2}{3}\]
By canceling the like terms, we get,
\[\dfrac{r+1}{n-r}=\dfrac{2}{3}\]
By cross multiplying the above equation, we get,
\[3\left( r+1 \right)=2\left( n-r \right)\]
Or, \[3r+3=2n-2r\]
Or, \[2n-2r-3r=3\]
\[\Rightarrow 2n-5r=3\]
By multiplying the above equation by 2 on both sides, we get,
\[4n-10r=6....\left( vi \right)\]
Now by adding equations (v) and (vi), we get,
\[\Rightarrow \left( 10r-3n \right)+\left( 4n-10r \right)=3+6\]
Or, \[-3n+4n=9\]
Therefore we get n = 9.
Now, to find \[^{n}{{C}_{8}}\], we will substitute r = 8 in equation (i), we get,
\[^{n}{{C}_{8}}=\dfrac{n!}{8!\left( n-8 \right)!}\]
By substituting the value of n = 9 in the above equation, we get,
\[^{n}{{C}_{8}}={{\text{ }}^{9}}{{C}_{8}}=\dfrac{9!}{8!\left( 9-8 \right)!}\]
Here, we can write 9! = 9 x 8!. So, we get,
\[^{n}{{C}_{8}}={{\text{ }}^{9}}{{C}_{8}}=\dfrac{9\times 8!}{8!1!}\]
By canceling the like terms, we get,
\[^{n}{{C}_{8}}={{\text{ }}^{9}}{{C}_{8}}=9\]
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 \[^{n}{{C}_{r}}\] that is \[\dfrac{n!}{r!\left( n-r \right)!}\] as it is a very useful formula for mathematics.