There are 3 piles of books on the table you need to arrange them in the self such that 1st book of each pile gets placed on the self,what is the possible no of arrangements:
a)12! b)12C3 * 9C4 c)12^12 d)none

3 piles i.e.  we have to select 1 book from each piles,

 ³C₁*³C₁*³C₁

=( 3!/1!*2! ) 3

 =3*3

=9   and hence answer is (D)

There can be 4 powers of 7 which will give the last 4 digits as :

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

Other powers will be of the form: 4k, 4k+1, 4k+2, 4k+3. And all the four have an equal probability of being selected.

So, total number of combinations = 4x4 (repetition being allowed)

Total number of favourable combinations = (7,3)(3,7)(9,1)(1,9) - 4

So probability = 4/16 = 1/4