Maharashtra State BoardHSC Commerce 11th
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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? - Mathematics and Statistics

Sum

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?

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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

Concept: Permutations When All Objects Are Not Distinct
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 6 Permutations and Combinations
Exercise 6.4 | Q 10 | Page 83
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