Question

16. How many different words can be formed with the letters of the word 'RAJARAM'? In how many of

Answer 1

Answer:

The word "ARRANGE" can be arranged in

2!×2!

7!

=

4

5040

=1260 ways.

For the two R's do occur together, let us make a group of R's taking from "ARRANGE" and permute them.

Then the number of ways =

2!

6!

=360.

The number ways to arrange "ARRANGE", where two "R's" will not occur together is =1260−360=900.

Also in the same way, the number of ways where two "A's" are together is 360.

The number of ways where two "A's" and two "R's" are together is 5!=120.

The number of ways where neither two "A's" nor two "R's" are together is =1260−(360+360)+120=660.

Step-by-step explanation:

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Answer 2

(i) Take RRJ out and the rest letter is AAAM and permutation wise it should form 4! variations, may the 3As being Afirst, Asecond, and Athird. These three alone may form 3! permutations and when setting Afirst = Asecond =Athird one would have 3! sets of identical variations so 4!/3! = 4. And intuitively we can already see them: AAAM, AAMA, AMAA and MAAA. While inserting anything (in our case RRJ) back into a 4 letter queue, there are exactly 4+1 spots to insert (before every letter + behind the last letter) so for RRJ there would be 4*(4+1) variations which is 20. If considering RJR or JRR also applies to the answer, 20+20+20=60.