Question

How many permutations of 4 letters can be made out of the letters of the word ‘examination’?

Answer 1

We have 3 cases here. All the letters in the four letter word are distinct, one letter repeats itself two times and two letters repeat themselves two times. We use the formula for selection of $r$ objects out of $n$ distinct objects as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}=\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r\left( r-1 \right)...1}$, the permutation or arrangement of $n$ distinct objects as $n!$ and the permutation of $n$objects where $m$ objects repeats themselves by ${{p}_{1}},{{p}_{2}},...,{{p}_{m}}$ times as $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!...{{p}_{m}}!}$ to get number of words from each case which we add to get the total number of words. $$$$

Complete step-by-step solution:
We are given the word in the question ‘examination’. We observe the word ‘examination’ and find that it has repeating letters a, i, n which repeat themselves 2 times each. There are 5 distinct letters e, x, m, t, o. If we are going to make 4 letter words we have 4 positions to fill.

\\_ & \\_ & \\_ & \\_ \\\ 1 & 2 & 3 & 4 \\\ \end{matrix}$$ We see that we can fill the four positions in one of three ways. $$$$ Case-1: We fill 4 positions such that no letter is repeated. So we can select 4 letters out 8 letters (a, i, n, e, x, m, t, o) in ${}^{8}{{C}_{4}}$ ways and then fill them in the positions. We show one such possible word below $$\begin{matrix} \underline{a} & \underline{t} & \underline{n} & \underline{m} \\\ 1 & 2 & 3 & 4 \\\ \end{matrix}$$ We can arrange the filled letters in $4!$ ways. .We use rule of product and find number of words from case-1 as $${{N}_{1}}={}^{8}{{C}_{4}}\times 4!=\dfrac{8!}{4!4!}\times 4!=5\times 6\times 7\times 8=1680$$ Case-2: We fill 4 positions such that no letter is repeated with two positions filled with 1 repeated letters. We show one such possible word below $$\begin{matrix} \underline{a} & \underline{t} & \underline{a} & \underline{m} \\\ 1 & 2 & 3 & 4 \\\ \end{matrix}$$ We can choose 1 repeated letter out 3 repeated letters (a , i , n) in ${}^{3}{{C}_{1}}$ ways and fill two positions out of 4. We can fill the rest 2 positions by selecting 2 distinct letters out of 7 letters 5 distinct e, x, m, t, o in ${}^{7}{{C}_{2}}$ ways. So we have selected $n=4$ letters out which one letter has to repeat itself by ${{p}_{1}}=2$ times which we can permute in $\dfrac{n!}{{{p}_{1}}!}=\dfrac{4!}{2!}$ ways. We use the rule of product and find the number of words from case-2 as $${{N}_{2}}=\dfrac{{}^{3}{{C}_{1}}\times {}^{7}{{C}_{2}}\times 4!}{2!}=3\times \dfrac{7\times 6}{1\times 2}\times 3\times 4=756$$ Case-3: We fill the four positions with 2 repeated letters. We can select 1 repeated letter out 3 repeated letters (a , i , n) in ${}^{3}{{C}_{2}}$ ways and fill 4 positions.We show one such possible word below $$\begin{matrix} \underline{a} & \underline{n} & \underline{a} & \underline{n} \\\ 1 & 2 & 3 & 4 \\\ \end{matrix}$$ So we have selected $n=4$ letters where two letters have to repeat themselves by ${{p}_{1}}={{p}_{2}}=2$ times which we can permute in $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!}=\dfrac{4!}{2!2!}$ ways. We use the rule of product and find number of words from case-3 as $${{N}_{3}}={}^{3}{{C}_{2}}\times \dfrac{4!}{2!2!}=\dfrac{3\times 2}{2\times 1}\times 3\times 2=18$$ We use rule of sum find permutations of 4 letters that can be made out of the letters of the word ‘examination’$N$ that can be made as the sum of number words from case-1, case-2, case-3 as $$N={{N}_{1}}+{{N}_{2}}+{{N}_{3}}=1680+756+18=2454$$ **Note:** We note that if we can do one thing in $m$ ways and another thing in$n$ ways then by rule of sum we can do either of them in $m+n$ ways and by the rule of the product we can do both of them in $mn$ ways. We note that the positions are fixed here if they would not have we would have considered derangement ${{D}_{n}}=n!\times \sum\limits_{i=0}^{n}{\dfrac{{{\left( -1 \right)}^{n}}}{i!}}$ .