Question

is there any body to slove it . .?& capacity to explain it?..
Now show ur talent,,,,;;;!!!!!

Answer 1

The word Mahabharat contains 10 letters.

It contains 4A's, 2H's, 1B, 1T, 1M, 1R.

The required number of ways = 10!/4!2!1!1!1!

                                                   = 10!/4!2!.



(1) Given that B and T are never together.

Consider That as 1-unit(10 - 1)! = 9!.

= 9!/4!2!1!1!

= 9!/4!2!.


Therefore the required number of ways = 10!/4!2! - 9!/4!2!

                                                                    = 3265920/48

                                                                    = 68040.



(2) M,A,T,H occurs at first four places.:


Observe this carefully:


-     -   -    - 

M  A   T   H   ---------- Now, these four letters can be arranged themselves in4! ways.



The remaining letters can be arranged in = A,B,H,A,R,A

                                                                      = 6!/3!  ----- Here 3! = (3 A's)




Therefore the total number of ways = 6!/3! * 4!

                                                             = 4 * 5 * 6 * 4 * 3 * 2 * 1

                                                             = 2880.




Hope this helps!

Answer 2

Hi,

Please see the attached file !



Thanks