Question

Five-letter words are to be formed out of the letters of the word INFINITESIMAL. What is their number?

Answer 1

Hint: For finding out the number of various combinations of words we must use selection principle and allot alphabets to respective positions to form a five-letter word. Also, the number of mutually distinguishable permutations of n things, taken all at a time, of which p are alike of one kind, q alike of second such that p + q = n is,$\dfrac{n!}{p!\cdot q!}$.

Complete step-by-step answer:
Now, we have to consider all cases of forming five letters from all letters different to letters alike where it is possible.
In the word INFINITESIMAL, the total occurrence of each letter can be stated as 4I, 2N, all other letters different. In total there are nine different types of letters.
Total number of alphabets = 9
Number of letters taken = 5
For evaluation of ${}^{n}{{C}_{r}}$, we use the formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
First, for all different alphabets the number of five-letter words:
Number of ways $={}^{n}{{C}_{r}}\cdot r!$
\[\begin{align}
  & {}^{9}{{C}_{5}}\cdot 5!=\dfrac{9!}{5!4!}\cdot 5! \\
 & =\dfrac{9\times 8\times 7\times 6\times 5\times 4!\times 5!}{4!\times 5!} \\
 & =9\times 8\times 7\times 6\times 5 \\
 & =15120 \\
\end{align}\]
Second case, For two alike and three different:
Number of choosing two alike letters from I, N $={}^{n}{{C}_{r}}={}^{2}{{C}_{1}}$
Number of choosing three different letters from other alphabets $={}^{8}{{C}_{3}}$
So, the number of ways $={}^{2}{{C}_{1}}\cdot {}^{8}{{C}_{3}}\cdot \dfrac{5!}{2!}$
$\begin{align}
  & =\dfrac{2!}{1!\times 1!}\times \dfrac{8!}{3!\times 5!}\times \dfrac{5!}{2!} \\
 & =\dfrac{8\times 7\times 6\times 5\times 4\times 3!\times 2!\times 5!}{3!\times 2!\times 5!} \\
 & =8\times 7\times 6\times 5\times 4 \\
 & =6720 \\
\end{align}$
Third case, For two alike, two alike and one different:
Number of choosing two alike letters $={}^{n}{{C}_{r}}={}^{2}{{C}_{2}}$
Number of choosing one different letter from other alphabets $={}^{7}{{C}_{1}}$
So, the number of ways $={}^{2}{{C}_{2}}\cdot {}^{7}{{C}_{1}}\cdot \dfrac{5!}{2!2!}$
$\begin{align}
  & =\dfrac{2!}{2!\times 0!}\times \dfrac{7!}{1!\times 6!}\times \dfrac{5!}{2!\times 2!} \\
 & =\dfrac{7\times 6!\times 5\times 4\times 3\times 2!\times 2!}{6!\times 2!\times 2!\times 2!} \\
 & =7\times 5\times 3\times 2 \\
 & =210 \\
\end{align}$
Fourth case, For three alike, two different:
Number of choosing three alike letters $={}^{1}{{C}_{1}}$
Number of choosing one different letter from other alphabets $={}^{8}{{C}_{2}}$
So, the number of ways $={}^{1}{{C}_{1}}\cdot {}^{8}{{C}_{2}}\cdot \dfrac{5!}{3!}$
$\begin{align}
  & =\dfrac{1!}{1!\times 0!}\times \dfrac{8!}{2!\times 6!}\times \dfrac{5!}{3!} \\
 & =\dfrac{8\times 7\times 6!\times 5\times 4\times 3!}{6!\times 3!\times 2!} \\
 & =8\times 7\times 5\times 2 \\
 & =560 \\
\end{align}$
Fifth case, For three alike, two alike:
Number of choosing three alike letters $={}^{1}{{C}_{1}}$
Number of choosing two alike letters $={}^{1}{{C}_{1}}$
So, the number of ways $={}^{1}{{C}_{1}}\cdot {}^{1}{{C}_{1}}\cdot \dfrac{5!}{3!2!}$
$\begin{align}
  & =\dfrac{1!}{1!\times 0!}\times \dfrac{1!}{1!\times 0!}\times \dfrac{5!}{3!\times 2!} \\
 & =\dfrac{5\times 4\times 3!}{3!\times 2!} \\
 & =5\times 2 \\
 & =10 \\
\end{align}$
Final case, For four alike, one different:
Number of choosing four alike letters $={}^{1}{{C}_{1}}$
Number of choosing one different letter from other alphabets $={}^{8}{{C}_{1}}$
So, the number of ways $={}^{1}{{C}_{1}}\cdot {}^{8}{{C}_{1}}\cdot \dfrac{5!}{4!}$
$\begin{align}
  & =\dfrac{1!}{1!\times 0!}\times \dfrac{8!}{1!\times 7!}\times \dfrac{5!}{4!} \\
 & =\dfrac{8\times 7!\times 5\times 4!}{7!\times 4!} \\
 & =8\times 5 \\
 & =40 \\
\end{align}$
Therefore, the total sum of all the cases $=15120+6720+210+560+10+40$
Total number $=22660$.

Note: The key concept involved in solving this problem is the knowledge of permutation under certain restrictions. Students must be careful while forming the possible cases of arrangements because the most common mistake is the formation of less number of cases than the total possible cases.