### NBHM 2020 PART A Question 8 Solution

You are given $20$ identical balls and $5$ bins that are colored differently (so that any two of the bins can be distinguished from each other). In how many ways can the balls be distributed into the bins in such a way that each bin has at least two balls?
Solution:
First, take $10$ balls in the hand. Consider the $5$ bins and put $2$ balls in each of these bins so that each bin will have at least $2$ balls. Since the balls are identical this can be done in only one way. Now, we are left with $10$ identical balls which have to be distributed in $5$ bins without any constraints. We know that the number of ways to put $n$ identical objects into $k$ labeled bins is equal to the binomial coefficient $$\binom{n+k-1}{k-1}$$ by stars and bars (CHECK THIS LINK). In our case, this number is equal to $$\binom{10+5-1}{5-1} = 1001$$.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...