Question

How many seven letter words can be formed without repetition out of the letters of the word MISTAKE such that the vowels go only into the odd places.
A.180
B.786
C.625
D.576

Answer 1

Hint: To solve the problem we use method of permutation method. Permutation is an arrangement that can be made by taking some or all of a number of given things. To find number of permutations of n different things taken r at a time we use
\[^n{P_r} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  {n - r} \,}} \right. }}\]
where n is n different things, r is number of things taken at a time and$\left| \!{\underline {\,
  {} \,}} \right. $is a factorial sign.

Complete step-by-step answer:
Now there are seven letter words MISTAKE. Out of which there are 3 vowels that are I, A and E. and there are four consonants that are M, S, T and K. let us mark out the position to be filled as follows
 \[(\dfrac{1}{x}).(2).(\dfrac{3}{x}).(4).(\dfrac{5}{x}).(6).(\dfrac{7}{x})\]
Because, vowels occupy odd positions therefore three vowels can be arranged in four cross (x). so, the number of ways of arranging the vowels is $^4{P_3}$.
$^4{P_3} = \dfrac{{4 \times 3 \times 2 \times 1}}{1} = 24$
Also, there are 4 consonants at the remaining four positions. These consonants can be arranged in $^4{P_4}$ ways.
$^4{P_4} = \dfrac{{\left| \!{\underline {\,
  4 \,}} \right. }}{{\left| \!{\underline {\,
  {4 - 4} \,}} \right. }} = \dfrac{{4 \times 3 \times 2 \times 1}}{{0!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{1} = 24$
Therefore, the required number of ways the word mistake can be arranged. So that vowels occupy odd position is 24× 24 = 576
The correct option is D.

Note: In permutation, the first place can be filled up in a different way as any one of the n person can be placed there. After filling up the first place in one n ways. There are (n-1) different ways to filing up second place . Similarly the third place can be filled by (n-2) different ways.
Proceeding in this ways, we see that
Whenever a place is filled up a new factor is introduced.
The factor begins with n and goes on diminishing by unity. therefore rth factor = n-(r-1)
Therefore , number of ways filling up r places = n(n-1)(n-2) …………………(n-r+1)
∴ $^n{P_r} = \dfrac{{n.(n - 1).(n - 2)........(n - r + 1)......3.2.1}}{{(n - r).......3.2.1}}$
∴ \[^n{P_r} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  {n - r} \,}} \right. }}\]