Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements the two R’s and two A’s are not together?

#### Solution

The word ARRANGE has 7 letters of which A and R are repeated 2 times.

∴ The number of ways of arranging letters of the word

= `(7!)/(2!2!)`

= `((7 xx 6 xx 5 xx 4 xx 3 xx 2)!)/(2!2!)`

= 1260

Here, we have to find the number of arrangements in which two R’s nor A’s are together.

A: set of words having 2A together

B: set of words having 2R together

Number of words having both A and both R not together

= 1260 − n(A ∪ B)

= 1260 − [n(a) + n(B) − n(A ∩ B)] .......(i)

n(A) = number of ways in which (AA) R, R, N, G, E are to be arranged

∴ n(A) = `(6!)/(2!)` = 360

n(B) = number of ways in which (RR), A, A, N, G, E are to be arranged

∴ n(B) = `(6!)/(2!)` = 360

n(A ∩ B) = number of ways in which (AA), (RR), N, G, E are to be arranged

∴ n(A ∩ B) = 5! = 120

Substituting n(A), n(B), n(A ∩ B) in (i), we get

Number of words having both A and both R not together

= 1260 – [360 + 360 – 120]

= 1260 − 600

= 660