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Question: If \(^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5\) , then find the value of \(r\) ?...

If 15Cr:15Cr−1=11:5^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5 , then find the value of rr ?

Explanation

Solution

For answering this question we will use the formulae for expansion of combinations that is nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} and simplify the ratio which we have from the question 15Cr:15Cr−1=11:5^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5 and derive the value of rr . We should remember that the rational numbers can never be assigned in combinations or permutations.

Complete step-by-step answer:
Now considering from the question we have the ratio given as 15Cr:15Cr−1=11:5^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5 .
From the basic concept we know the formulae for expansion of combinations that is nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . After using this we will have 15!r!(15−r)!:15!(r−1)!(15−(r−1))!=11:5\dfrac{15!}{r!\left( 15-r \right)!}:\dfrac{15!}{\left( r-1 \right)!\left( 15-\left( r-1 \right) \right)!}=11:5 .
From the basic concept we know that the rational numbers can never be assigned in combinations or permutations.
After further simplifying this we will have 1r.(r−1)!(15−r)!:1(r−1)!(16−r)!=11:5\dfrac{1}{r.\left( r-1 \right)!\left( 15-r \right)!}:\dfrac{1}{\left( r-1 \right)!\left( 16-r \right)!}=11:5 .
By simplifying this we will have (r−1)!(16−r)!r.(r−1)!(15−r)!=11:5\dfrac{\left( r-1 \right)!\left( 16-r \right)!}{r.\left( r-1 \right)!\left( 15-r \right)!}=11:5 .
Now we will derive the simplified expression
(16−r).(15−r)!r.(15−r)!=11:5 ⇒(16−r)r=11:5 \begin{aligned} & \dfrac{\left( 16-r \right).\left( 15-r \right)!}{r.\left( 15-r \right)!}=11:5 \\\ & \Rightarrow \dfrac{\left( 16-r \right)}{r}=11:5 \\\ \end{aligned} .
Hence from (16−r)r=11:5\dfrac{\left( 16-r \right)}{r}=11:5 we can conclude that the value of rr is 55 .

Note: While answering questions of this type we should be sure with the calculations and formulae if by mistake we had used the formulae of permutations which is nCr=n!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!} instead of combinations formulae which is nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} then we will have 15!(15−r)!:15!(15−(r−1))!=11:5\dfrac{15!}{\left( 15-r \right)!}:\dfrac{15!}{\left( 15-\left( r-1 \right) \right)!}=11:5 . Make a note that the rational numbers can never be assigned in combinations or permutations. After simplifying this we will have 1(15−r)!:1(16−r)!=11:5\dfrac{1}{\left( 15-r \right)!}:\dfrac{1}{\left( 16-r \right)!}=11:5 . By further simplifying we will have (16−r)=115\left( 16-r \right)=\dfrac{11}{5} . Here we will end up having r=695r=\dfrac{69}{5} which is not possible since rational numbers are not allowed. Hence we should be careful.