Question

# Consider the letters of the word ′ARRANGE′. Then the possible number of words

A
when 2Rs are not together is 5!2!× 6P2=900
B
when two As are together but not two R's is 240
C
when the vowels are together is 5!3!2!2!
D
when both As are together and both Rs are together is 120

Solution

## The correct options are B when two A′s are together but not two R's is 240 C when the vowels are together is 5!3!2!2! D when both A′s are together and both R′s are together is 120When 2R′s are not together. Arrange the letters ′A,A,N,G,E in alternate manner, such that there will be 6 spaces left, so that can be filled by 2R′s in 6C2 ways So the total possible words =5!2!× 6C2=900 When two A′s are together but not two R′s. The number of words in which both A's are together  =6!2! [consider both the A′s as one unit] =360 The number of words in which both A′s and both R′s are together is 5!=120 [consider both the A′s as one unit and both the R′s as one unit] Therefore, the number of words in which both A′s are together but the two R′s are not together is 360−120=240 When the vowels are together. Make one group of all vowels. The total ways of arrangement is =5!×3!2!2! When both A′s are together and both R′s are together. Make one group 2A′s and another group of 2R′s. The total ways of arrangement is =5!=120Mathematics

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