Question

If n is an integer with $0 \leqslant n \leqslant 11$, then the minimum value of $n!\left( {11 - n} \right)!$ is attained when a value of $n$ is equal to?
1)$$$11$ 2)5$ $$3)7 $$4)$$$9

Answer 1

We have to find the value of $n$ for which we have the minimum value of $n!\left( {11 - n} \right)!$ . We solve this using the concept of permutation and combination . We should have the knowledge of formulas for combinations of terms . And how to expand the terms of a factorial . We will simplify the formula in terms of combination and put the values in the formula to find the min value of $n!\left( {11 - n} \right)!$.

Complete step-by-step solution:
Given :
$0 \leqslant n \leqslant 11$
We have to find the minimum value of $n!\left( {11 - n} \right)!$
We know that ,
Formula of combination is given as :
${}^r{C_n} = \dfrac{{r!}}{{\left[ {n! \times \left( {r - n} \right)!} \right]}}$
The value of $r$ is $11$ .
Using the formula of combination we can write the expression as ,
${}^{11}{C_n} = \dfrac{{11!}}{{\left[ {n! \times \left( {11 - n} \right)!} \right]}}$
Now , we have to find the minimum value of $n!\left( {11 - n} \right)!$. This means that for finding the minimum value of $n!\left( {11 - n} \right)!$ The value of ${}^{11}{C_n}$ would be maximum .
The maximum value of ${}^{11}{C_n}$ should be at $\;n = 6$ or $n = 5$
( As for any other value of n the value of factorial in the denominator of expansion of ${}^{11}{C_n}$ would increase which decreases the value of ${}^{11}{C_n}$ )
Hence ,
Put$n = 5$, we get
${}^{11}{C_5} = \dfrac{{11!}}{{\left[ {5! \times \left( {11 - 5} \right)!} \right]}}$
${}^{11}{C_5} = \dfrac{{11!}}{{\left[ {5! \times 6!} \right]}}$
Put$n = 6$, we get
${}^{11}{C_6} = \dfrac{{11!}}{{\left[ {6! \times \left( {11 - 6} \right)!} \right]}}$
${}^{11}{C_6} = \dfrac{{11!}}{{\left[ {6! \times 5!} \right]}}$
Thus the value of ${}^{11}{C_n}$ is equal for $n = 5$ and $\;n = 6$
Thus the value of n for which $n!\left( {11 - n} \right)!$ is $5$ .
Hence , the correct option is $\left( 2 \right)$.

Note: Corresponding to each combination of ${}^n{C_r}$ we have $r!$ permutations, because $r$ objects in every combination can be rearranged in $r!$ ways . Hence , the total number of permutations of n different things taken $r$ at a time is ${}^n{C_r} \times r!$. Thus $\;{}^n{P_r} = {}^n{C_r} \times r!,0 < r \leqslant n$
Also , some formulas used :
${}^n{C_1} = n$
${}^n{C_2} = \dfrac{{n\left( {n - 1} \right)}}{2}$
${}^n{C_0} = 1$
${}^n{C_n} = 1$