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Permutation under Restriction

How many diff...

Question

How many different words can be formed by arranging letters of work 'ARRANGE' if neither two R's nor two A's occur together?

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Solution

Total number of arrangements possible $= \frac{7!}{2! 2!} = 1260$
Total number of arrangements in which 2R's are together $= \frac{6!}{2!} = 360$
Total number of arrangements in which 2A's are together $= \frac{6!}{2!} = 360$
Total number of arrangements in which 2A's as well as 2R's are together $= 5! = 120$
Therefore total number of arrangements in which neither 2A's nor 2R's are together $= 1260 - 360 - 360 + 120 = 660$

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