Question

If \[{}^{n}{{C}_{r-1}}=36,{}^{n}{{C}_{r}}=84\ and\ {}^{n}{{C}_{r+1}}=126\] , then find the values of n and r by evaluating the three equations.

Answer 1

HINT: - The expansion of \[{}^{n}{{C}_{r}}\] is \[=\dfrac{n!}{r!\left( n-r \right)!}\] (Where n are r are the same numbers as in \[{}^{n}{{C}_{r}}\] )

Complete step-by-step answer:
As mentioned in the question, the first equation can be written using the formula mentioned in the as
\[{}^{n}{{C}_{r-1}}=\dfrac{n!}{(r-1)!\left( n-(r-1) \right)!}=36\ \ \ \ \ ...\left( a \right)\]
The second equation can be written using the formula mentioned in the hint as
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}=84\ \ \ \ \ ...\left( b \right)\]
Similarly, the third equation can be written using the help of the formula mentioned in the hint as
\[{}^{n}{{C}_{r+1}}=\dfrac{n!}{(r+1)!\left( n-(r+1) \right)!}=126\ \ \ \ \ ...(c)\]

Now, using the equations above mentioned that are (a), (b) and (c),
On dividing (a) and (b), we get
\[\dfrac{\dfrac{n!}{(r-1)!\left( n-(r-1) \right)!}}{\dfrac{n!}{r!\left( n-r \right)!}}=\dfrac{36}{84}\]
On simplifying the above equation, we get
\[\dfrac{r}{(n-r+1)}=\dfrac{3}{7}\]
On cross multiplying, we get
\[10r=3n+3\ \ \ \ ...(d)\]
Similarly, on dividing (b) and (c), we get
\[\dfrac{\dfrac{n!}{r!\left( n-r \right)!}}{\dfrac{n!}{(r+1)!\left( n-(r+1) \right)!}}=\dfrac{84}{126}\]
On simplifying the above equation, we get
\[\dfrac{\left( r+1 \right)}{(n-r)}=\dfrac{2}{3}\]
On cross multiplying, we get
\[5r=2n-3\ \ \ \ \ (e)\]
Now, to get the value or to find out the value of n and r that has been used in these equations, we can perform elimination method to get the values.
So,
On multiplying equation (e) with 2, we get
\[10r=4n-6\ \ \ \ \ ...(f)\]
Now, on equating equation (d) and (f), we get
\[\begin{align}
  & 3n+3=4n-6 \\
 & n=9 \\
\end{align}\]
Now, we have got the value of n, so we can just put the value of n in any of the equations and then get the value of r.
So, on putting the value of n in (f), we get
\[\begin{align}
  & 10r=4\times 9-6 \\
 & 10r=36-6 \\
 & 10r=30 \\
 & r=3 \\
\end{align}\]
Hence, the value of n = 9 and the value of r = 3. \[\]

NOTE:The students can make an error in evaluating the value of \[{}^{n}{{C}_{r}}\] and might confuse with \[{}^{n}{{P}_{r}}\] which has a similar expansion to \[{}^{n}{{C}_{r}}\] that is \[\dfrac{n!}{r!}\] which will give a wrong answer when the question is done using it.

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