Question

The number of five-letter words that are formed out of the letters of the word INFINITESIMAL.
A. 15120
B. 22660
C. 22450
D. 22440

Answer 1

Hint: The above case given evolves the concept of selection from different objects, selection from identical objects and selection when both identical and distinct objects are present. The word given out of which we need to form five letter words has a total 13 letters in it.

Complete step-by-step answer:
The 13 letters are 4I, 2N and S, M, A, L, F , T, E which are seven different letters or objects. We can conclude that there are a total 9 types of letters and objects. Now we have 9 types of objects to fill 5 places as we have to make 5 letter words. Therefore we will have different cases for all the arrangements of all the possible selections.
When all the letters in 5 letter words are different that is all the places will have different letters, it is arranged as
$^9{P_5} = \dfrac{{\left| \!{\underline {\,
  9 \,}} \right. }}{{\left| \!{\underline {\,
  {9 - 5} \,}} \right. }} = \dfrac{{\left| \!{\underline {\,
  9 \,}} \right. }}{{\left| \!{\underline {\,
  4 \,}} \right. }} = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times \left| \!{\underline {\,
  4 \,}} \right. }}{{\left| \!{\underline {\,
  4 \,}} \right. }} = 9 \times 8 \times 7 \times 6 \times 5 = 15120$
Second case is when the five places are filled with 2 letters of one kind and three completely different letters, therefore the similar letters can be arranged from 2 types either I or N and utilizing one of the types we will have eight different types for the remaining places to be filled.
$
  ^2{C_1}{.^8}{C_3}.\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  2 \,}} \right. }} = \dfrac{{\left| \!{\underline {\,
  2 \,}} \right. }}{{\left| \!{\underline {\,
  1 \,}} \right. \left| \!{\underline {\,
  {2 - 1} \,}} \right. }} \times \dfrac{{\left| \!{\underline {\,
  8 \,}} \right. }}{{\left| \!{\underline {\,
  {8 - 3} \,}} \right. .\left| \!{\underline {\,
  3 \,}} \right. }} \times \dfrac{{5 \times 4 \times 3 \times \left| \!{\underline {\,
  2 \,}} \right. }}{{\left| \!{\underline {\,
  2 \,}} \right. }} \\
   = 2 \times \dfrac{{\left| \!{\underline {\,
  8 \,}} \right. }}{{\left| \!{\underline {\,
  5 \,}} \right. .\left| \!{\underline {\,
  3 \,}} \right. }} \times 60 = 120 \times \dfrac{{8 \times 7 \times 6}}{6} = 120 \times 56 = 6720 \\
 $
Third case is when the five places are filled with 2 objects of one kind and 2 objects of other kind and the left one place with 1 different letter and we know that there are seven different letters.
$^2{C_2}{.^7}{C_1}.\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  2 \,}} \right. .\left| \!{\underline {\,
  2 \,}} \right. }} = 1.\dfrac{{\left| \!{\underline {\,
  7 \,}} \right. }}{{\left| \!{\underline {\,
  1 \,}} \right. .\left| \!{\underline {\,
  6 \,}} \right. }} \times \dfrac{{5 \times 4 \times 3}}{2} = 7 \times 30 = 210$
Fourth case is when the five places are filled with 3 letters of one kind and the left two places with 2 different letters.
$^1{C_1}{.^8}{C_2}.\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  3 \,}} \right. }} = 1.\dfrac{{\left| \!{\underline {\,
  8 \,}} \right. }}{{\left| \!{\underline {\,
  2 \,}} \right. .\left| \!{\underline {\,
  6 \,}} \right. }}.20 = \dfrac{{8 \times 7}}{2} \times 20 = 560$
Fifth case is when 3 letters are one of a kind and other two letters of other kind which will be from 4I and 2N respectively.
$^1{C_1}{.^1}{C_1}.\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  3 \,}} \right. .\left| \!{\underline {\,
  2 \,}} \right. }} = 1 \times 1 \times \dfrac{{5 \times 4}}{2} = 10$
Sixth case is when four letters are same and the left one place is filled by a different letter
$^8{C_1}{.^1}{C_1}.\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  4 \,}} \right. }} = \dfrac{{\left| \!{\underline {\,
  8 \,}} \right. }}{{\left| \!{\underline {\,
  7 \,}} \right. .1}} = 8 \times 5 = 40$
Adding all the above we get total arrangements,
$ 15120+6720+210+560+10+40=22660 $
So, the correct answer is “Option B”.

Note: Therefore the permutation of n objects taken all at a time when p objects are alike and of one kind, q objects are alike and of another kind, r objects are alike of yet another kind and the remaining objects are different is
$\dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  p \,}} \right. \left| \!{\underline {\,
  q \,}} \right. \left| \!{\underline {\,
  r \,}} \right. }}$ .