In how many ways can we put $31$ people in $3$ rooms such that each room has an odd number of people ?
Can I apply generating functions here or Sterling's formula of $2$nd kind or any other method ?
www.math.toronto.edu/szegedy/solutions.pdf
The above link gives an approach, but I am not familiar with it .
@trueblueannil demonstrated that my original solution was incorrect since I had overlooked the fact that arrangements within each room did not matter. I have edited my solution to fix the problem.
Preliminary remarks: The Binomial Theorem states that $$(x + y)^n = \sum_{k = 0}^{n} \binom{n}{k}x^{n - k}y^k$$ Substituting $1$ for both $x$ and $y$ yields $$2^n = (1 + 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}1^k = \sum_{k = 0}^n \binom{n}{k}$$ Since $\binom{n}{k}$ is the number of subsets of size $k$ and a subset of a set with $n$ elements has up to $n$ elements, the preceding statement says that a set with $n$ elements has $2^n$ subsets. If we instead set $x = 1$ and $y = -1$ in the Binomial Theorem, we obtain $$0 = 0^n = (1 + (-1))^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}(-1)^k = \sum_{k = 0}^{n} \binom{n}{k}(-1)^k$$ from which it follows that the number of subsets of a set with $n$ elements that have an odd number of elements (the negative terms) is equal to the number of subsets with an even number of elements (the positive terms).
Solution: There are $\binom{31}{1}$ ways to select one person to be placed in the first room. Choosing who among the remaining people is placed in the second room determines who goes in the third room. Since an odd number of people must be placed in each room, we can select $1$, $3$, $5$, $7$, $9$, $11$, $13$, $15$, $17$, $19$, $21$, $23$, $25$, $27$, or $29$ of the remaining $30$ people to be placed in the second room. Hence, the number of arrangements in which one person is placed in the first room is $$\binom{31}{1}\left[\binom{30}{1} + \binom{30}{3} + \cdots + \binom{30}{27} + \binom{30}{29}\right] = \binom{31}{1}2^{30 - 1} = \binom{31}{1}2^{29}$$
There are $\binom{31}{3}$ ways to select three of the $31$ people to be placed in the first room. Since the number placed in each room must be odd, the number of people who can be placed in the second room is $1$, $3$, $5$, $7$, $9$, $11$, $13$, $15$, $17$, $19$, $21$, $23$, $25$, or $27$. Hence, the number of arrangements in which exactly three people are placed in the first room is $$\binom{31}{3}\left[\binom{28}{1} + \binom{28}{3} + \cdots + \binom{28}{25} + \binom{28}{27}\right] = \binom{31}{3}2^{28 - 1} = \binom{31}{3}2^{27}$$
Continuing in this way, we obtain the number of arrangements in which an odd number of the $31$ people are placed in each room, which is $$\binom{31}{2}2^{29} + \binom{31}{3}2^{27} + \binom{31}{5}2^{25} + \binom{31}{7}2^{23} + \binom{31}{9}2^{21} + \binom{31}{11}2^{19} + \binom{31}{13}2^{17} + \binom{31}{15}2^{15} + \binom{31}{17}2^{13} + \binom{31}{19}2^{11} + \binom{31}{21}2^9 + \binom{31}{23}2^7 + \binom{31}{25}2^5 + \binom{31}{27}2^3 + \binom{31}{29}2^1$$ $$= 154,418,349,070,986$$
Note: If we were dealing with indistinguishable objects, we could apply the following argument.
Let $x_1$ be the number of people placed in the first room; let $x_2$ be the number of people placed in the second room; let $x_3$ be the number of people placed in the third room. Then $$x_1 + x_2 + x_3 = 31 \tag{1}$$ Since each $x_k$ is a positive odd integer, if we set $x_k = 2y_k + 1$ for $1 \leq k \leq 3$, then each $y_k$ is a nonnegative integer, and \begin{align*} 2y_1 + 1 + 2y_2 + 1 + 2y_3 + 1 & = 31\\ 2y_1 + 2y_2 + 2y_3 & = 28\\ y_1 + y_2 + y_3 & = 14 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers with the same number of solutions that equation 1 has in the positive odd integers. The number of solutions of equation 2 can be found using the formula for combinations with repetition. However, people are distinguishable.