Class 11

Math

Algebra

Permutations and Combinations

Different words are being formed by arranging the letters of the word ARRANGE. All the words obtained are written in the form of a dictionary

Number of ways (words) in which neither two A's nor two R's comes together- $240$
- $360$
- $660$
- None of these

Among these, let us find the cases when the 2 R's are together.

So, leaving the 2 A's, we have: RR, N, G, E i.e. 4 objects to be arranged in $4!=24$ ways.

There are 5 gaps (including the ones at the ends) for placing the 2 A's which can be done in $_{2}C=10$ ways.

Thus, total cases where A's are not together while R's are together = $24×10=240$

Thus, cases where A's are not together and R's are not together = $900−240=660$

Hence, (c) is correct.