Question

Show that the number of six-letter words that can be formed using the letters of the word "assist" in which s’s alternate with other letters is 12. $$$$

Answer 1

We consider 6 empty places ${{P}_{1}},{{P}_{2}},{{P}_{3}},{{P}_{4}},{{P}_{5}},{{P}_{6}}$ where we can fill the letters from the word "assist". We have two cases here either we can fill ${{P}_{1}},{{P}_{3}},{{P}_{5}}$ with letter ‘s’ and fill the rest 3 distinct letters a, i, t and then permute them or we can fill ${{P}_{2}},{{P}_{4}},{{P}_{6}}$ with letter ‘s’ and fill the rest 3 distinct letters a, i, t and then permute them. We add the number of words from case-1 and case-2 and to find the total number of words 12.$$$$

Complete step-by-step solution:
We also know from permutation that the number of ways we can arrange $n$ distinct objects into a particular order is given by $n!$. $We are asked to prove the statement “The number of six letter words that can be formed using the letters of the word "assist" in which s’s alternate with other letters is 12.” We see that the word ‘assist’ contains 6 letters a, s, s, i, s t where s is repeated 3 times and there are 3 distinct letters a, i and t.$
Let us consider 6 empty places ${{P}_{1}},{{P}_{2}},{{P}_{3}},{{P}_{4}},{{P}_{5}},{{P}_{6}}$.$$$$
Case-1: We can fill the letter ‘s’ alternatively in the slot ${{P}_{1}},{{P}_{3}},{{P}_{5}}$ in only 1 way and we can place the rest 3 distinct letters a, i and t in the slots ${{P}_{2}},{{P}_{4}},{{P}_{6}}$ and arrange them in $3!$ ways. So by the rule of product the number of words in Case-1 is
${{n}_{1}}=1\times 3!=6$

{{P}_{1}} & {{P}_{2}} & {{P}_{3}} & {{P}_{4}} & {{P}_{5}} & {{P}_{6}} \\\ \underline{s} & \\_ & \underline{s} & \\_ & \underline{s} & \\_ \\\ \end{matrix}$$ Case-2: We can fill the letter ‘s’ alternatively in the slot ${{P}_{2}},{{P}_{4}},{{P}_{6}}$ in only 1 way and we can place the rest 3 distinct letters a, i and t in in the slots ${{P}_{1}},{{P}_{3}},{{P}_{5}}$ and arrange them in $3!$ ways. So by the rule of product the number of words in Case-2 is $${{n}_{2}}=1\times 3!=6$$ $$\begin{matrix} {{P}_{1}} & {{P}_{2}} & {{P}_{3}} & {{P}_{4}} & {{P}_{5}} & {{P}_{6}} \\\ \\_ & \underline{s} & \\_ & \underline{s} & \\_ & \underline{s} \\\ \end{matrix}$$ So by rule of sum, the total number of six letter words that can be formed using the letters of the word "assist" in which s’s alternate with other letters is sum of word i case-1 and case-2 that is $${{n}_{1}}+{{n}_{2}}=6+6=12$$ Hence the statement is proved. **Note:** We note to be careful that after filling up the letter ‘s’ in the slots we can arrange the three s’s in 1 way not in $3!$ ways. We note that the slots are fixed to fill the letters, if they would not have been given problems would have been a derangement problem. We can find the total number of 6 letter words using the letter of ‘assist’ by taking ‘ss’ as a single unit and find it as $\left( 6-1 \right)!=5!$.