Question

Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have letters R and H never together?

Answer 1

In the word 'MAHARASHTRA' the number of letters is n = 11 of which A repeats 4 times, i.e., p = 4, H repeats twice i.e., q = 2, R repeats twice, i.e., r = 2 and rest are distinct.

∴ the number of different arrangements with the letters of the word MAHARASHTRA is 

`("n"!)/("p"!"q"!"r"!)`

= `(11!)/(4!2!2!)`

= `(11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!)/(4! × 2 × 1 × 2 × 1)`

= 11 × 10 × 9 × 2 × 7 × 6 × 5

= 415800

When two R and two H are together forms one unit (object) and taking remaining 7 letters, we have total number of letters is n = 8 of which A repeats 4 times, i.e., p = 4.

Now both R and both H can be arranged together in `(4!)/(2!2!)`

∴ the total number of possible arrangements of letters in which both R and both H are together

= `(8!)/(4!) × (4!)/(2!2!)`

= `(8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1 × 2 × 1)`

= 8 × 7 × 6 × 5 × 3 × 2

= 10080

Hence, the number of arrangements of the letters in which R and H are never together = (Total number of arrangements) − (The number of arrangements in which R and H are together)

= 415800 − 10080

= 405720.