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

Hint: 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 :
\[{}^n{C_1} = n\]
\[{}^n{C_2} = \dfrac{{n\left( {n - 1} \right)}}{2}\]
\[{}^n{C_0} = 1\]
\[{}^n{C_n} = 1\]