Question

We have combinations \[^n{C_{r - 1}} = 36{,^n}{C_r} = 84\] and \[^n{C_{r + 1}} = 126\] then \[r\] is equal to
A). \[1\]
B). \[2\]
C). \[3\]
D). None of these

Answer 1

Hint: Here we are asked to find the value \[r\] from the given data. Here we will use the combination formula to find the value of \[r\]. First, we will expand given terms using a combination formula and form equations, and solve them to find the value \[r\].
Formula Used: Formula that we need to know:
\[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step-by-step solution:
It is given that \[^n{C_{r - 1}} = 36{,^n}{C_r} = 84\] and \[^n{C_{r + 1}} = 126\]. We aim to find the value of\[r\].
We know the formula for \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]. Using this let us expand the given terms.
\[^n{C_{r - 1}} = \dfrac{{n!}}{{r - 1!\left( {n - r + 1} \right)!}} = 36...........(1)\]
\[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} = 84...............(2)\]
\[^n{C_{r + 1}} = \dfrac{{n!}}{{r + 1!\left( {n - r - 1} \right)!}} = 126...........(3)\]
Now let us solve these equations to find the value of \[r\].
Dividing equation \[(2)\] by \[(1)\] we get
\[\dfrac{{n!}}{{r!\left( {n - r} \right)!}} \times \dfrac{{r - 1!\left( {n - r + 1} \right)!}}{{n!}} = \dfrac{{84}}{{36}}\]
Let us simplify the above expression.
\[\dfrac{{\left( {n - r + 1} \right)}}{r} = \dfrac{{84}}{{36}}\]
On further simplification we get
\[\dfrac{{\left( {n - r + 1} \right)}}{r} = \dfrac{{84}}{{36}}\]
\[\dfrac{{\left( {n - r + 1} \right)}}{r} = \dfrac{7}{3}\]
On cross multiplying the above, we get
\[3\left( {n - r + 1} \right) = 7r\]
\[\Rightarrow 3n - 3r + 3 = 7r\]
\[\Rightarrow 10r = 3n + 3..........(4)\]
Now let us divide the equation\[(3)\] by \[(2)\] we get
\[\dfrac{{n!}}{{r + 1!\left( {n - r - 1} \right)!}} \times \dfrac{{r!\left( {n - r} \right)!}}{{n!}} = \dfrac{{126}}{{84}}\]
On simplifying the above expression, we get
\[\dfrac{{n!}}{{\left( {r + 1} \right)r!\left( {n - r - 1} \right)!}} \times \dfrac{{r!\left( {n - r} \right)\left( {n - r - 1} \right)!}}{{n!}} = \dfrac{{126}}{{84}}\]
On further simplification we get
\[\dfrac{{n - r}}{{r + 1}} = \dfrac{3}{2}\]
On cross multiplying the above, we get
\[2\left( {n - r} \right) = 3\left( {r + 1} \right)\]
\[\Rightarrow 2n - 2r = 3r + 3\]
\[\Rightarrow 5r = 2n - 3.........(5)\]
Now let us divide the equation \[(4)\] by \[(5)\]we get
\[\dfrac{{10r}}{{5r}} = \dfrac{{3n + 3}}{{2n - 3}}\]
On simplifying the above expression, we get
\[2 = \dfrac{{3n + 3}}{{2n - 3}}\]
\[\Rightarrow 2\left( {2n - 3} \right) = 3n + 3\]
\[\Rightarrow 4n - 6 = 3n + 3\]
\[\Rightarrow n = 9\]
Now let us substitute the value \[n = 9\] the equation \[(4)\] we get
\[(4) \Rightarrow 10r = 3(9) + 3\]
On simplifying the above expression, we get
\[ \Rightarrow 10r = 30\]
\[ \Rightarrow r = 3\]
Thus, we got the value\[r = 3\]. Now let us see the options, option (a) \[1\] is an incorrect option as we got \[r = 3\] in our calculation.
Option (b) \[2\] is an incorrect option as we got \[r = 3\] in our calculation.
Option (c) \[3\] is the correct option as we got the same answer in the calculation above.
Option (d) None of these is an incorrect answer as we got option (c) as a correct option.
Hence, option (c) \[3\] is the correct answer.

Note: A factorial of any number \[n\] is written as \[n! = n(n - 1)(n - 2)(n - 3)...3.2.1\]. Thus, the factorial expansion can also be written as \[n! = n(n - 1)!\] since \[(n - 1)! = (n - 1)(n - 2)(n - 3)...3.2.1\]. We have used this form of expansion in our calculation above.