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
