Question

Out of 9 boys of a group n members is formed. What is the value of n so that the maximum number of groups can be formed?
a.4
b.5
c.4 or 5
d.6

Answer 1

Hint: Use the direct formula, first check the total boys are odd or even. If odd then, answer is $\dfrac{n+1}{2}$ or $\dfrac{n-1}{2}$ and if even, then answer is $\dfrac{n}{2}$.

Complete step-by-step answer:

In the question we are asked that out of 9 boys of n members are formed, so we have to find the value of n so that the maximum number of groups can be formed.
Now, if we have to choose ‘r’ people from ‘n’ people we can do it in ${}^{n}{{C}_{r}}$ ways and in the question we are asked for what value of r ${}^{n}{{C}_{r}}$ will be greatest.
Here, r will be in terms of ‘n’.
As we know that ${}^{n}{{C}_{k}}={}^{n}{{C}_{n-k}}$ so it suffices to tell that ${}^{n}{{C}_{k}}<{}^{n}{{C}_{r}}$ for $k${}^{n}{{C}{k}}=\dfrac{n\left( n-1 \right).....\left( n-k+1 \right)}{k!}$And${}^{n}{{C}{r}}=\dfrac{n\left( n-1 \right).....\left( n-r+1 \right)}{r!}$Now, we will divide${}^{n}{{C}{k}}$by${}^{n}{{C}{r}}$so, we get$\dfrac{\left( n-k \right)\left( n-k-1 \right).....\left( n-r+1 \right)}{r\left( r-1 \right).....\left( k+1 \right)}$……………………………………………(i) Now let’s note that$n-r+1\ge n-\dfrac{n+1}{2}+1$Which can be written as,$n-r+1\ge \dfrac{n+1}{2}$We know that,$r<\dfrac{n+1}{2}$So,$n-r+1>r$Hence,$r<\dfrac{n+1}{2}$Since the last factor of the numerator is greater than the first factor of denominator of fraction(i), so we have every factor of numerator greater than every factor of denominator so the fraction (i) is greater than 1. So for$n=even$$r=\left( \dfrac{n}{2} \right)$and for$n=odd$$r=\left( \dfrac{n+1}{2} \right)or\left( \dfrac{n-1}{2} \right)$As we know$n=9$so r is either$\dfrac{9+1}{2}or\dfrac{9-1}{2}$ which is 4 or 5.
So, the correct option is ‘C’.

Note: If the total number is given already and one has to find maximum number of groups so one can do by using fact that if $n=odd$ then r is either $\dfrac{n+1}{2}or\dfrac{n-1}{2}$ and if $n=even$ then r is $\dfrac{n}{2}$.