@vijju123 I am unable to understand how to solve this problem . I would be highly grateful to you if you could help me understand the above problem .

See first select the k marbles(1 from each),so now you are left with selecting n-k marbles.

This could be viewed as placing n-k marbles into k boxes.So the ans is the number of ways to distribute n-k indistinguishable objects in k distinguishable boxes which is (n-k+k-1)Choose(k-1)=(n-1)C(k-1)

(Basically formula for distributing r indist objects in k distinguishable box is (r+k-1)C(k-1) . U can find its proof easily ,if u google it

Thanks man !!

Can u please tell me why it is not (n-k+k-1)C(k) ?

Another way to see why the answer is \binom{n-1}{k-1} (or (n-1)C(k-1) ): imagine that you have n positions ready to fill with marbles, and to remind yourself to change to the next colour you can put a mark in between some of these positions. You will need k{-}1 marks for the colour changes that can go into some selection of the n{-}1 gaps in the ready positions.

Please illustrate your point with an example?