Question

If \[{}^{15}{{C}_{r}}:{}^{15}{{C}_{r-1}}=11:5\], find r.

Answer 1

Hint: The expression is in the form of \[{}^{n}{{C}_{r}}\]. Expand the formula for \[{}^{n}{{C}_{r}}\] and simplify the expression obtained and equate it to 11:5. Thus cross multiply and find r.

Complete step-by-step answer:

Given to us the expression, \[{}^{15}{{C}_{r}}:{}^{15}{{C}_{r-1}}=11:5\].
We know that this is of the form \[{}^{n}{{C}_{r}}\].
\[\therefore {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Now we can expand \[{}^{15}{{C}_{r}}:{}^{15}{{C}_{r-1}}\] similarly,
\[\dfrac{{}^{15}{{C}_{r}}}{{}^{15}{{C}_{r-1}}}=\dfrac{\dfrac{15!}{r!\left( 15-r \right)!}}{\dfrac{15!}{\left( r-1 \right)!\left( 15-r+1 \right)!}}\]
We cancel out 15! from numerator and denominator.
This is of the form \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}\], which we can write as, \[\dfrac{a}{b}\times \dfrac{d}{c}\].
\[\therefore \dfrac{15!}{r!\left( 15-r \right)!}\times \dfrac{\left( 15-r+1 \right)!\left( r-1 \right)!}{15!}=\dfrac{\left( 15-r+1 \right)!\left( r-1 \right)!}{\left( 15-r \right)!r!}\]
It is said that, \[{}^{15}{{C}_{r}}:{}^{15}{{C}_{r-1}}=11:5\].
\[\begin{align}
  & \Rightarrow \dfrac{\left( 15-r+1 \right)!\left( r-1 \right)!}{\left( 15-r \right)!r!}=\dfrac{11}{5} \\
 & \dfrac{\left( 15-r+1 \right)\left( 15-r \right)!\left( r-1 \right)!}{\left( 15-r \right)!r\times \left( r-1 \right)!}=\dfrac{11}{5} \\
\end{align}\]
Cancel (15 - r)! and (r - 1)! from the numerator and denominator.
\[\dfrac{\left( 15-r+1 \right)}{r}=\dfrac{11}{5}\]
Now let us cross multiply the above expression,
\[\begin{align}
  & 5\left( 15-r+1 \right)=11r \\
 & 75-5r+5=11r \\
 & \Rightarrow 11r+5r=80 \\
 & 16r=80 \\
 & \therefore r=\dfrac{80}{16}=5 \\
\end{align}\]
Thus we got the value of r = 5.

Note: The number of possibilities for choosing an ordered set of r objects. Thus to solve a particular problem we need to know the expression of \[{}^{n}{{C}_{r}}\].