The chairs

Each way to distribute the 100 chairs to the four rooms as described correspond to a $x^{10}$ term in the following product- $(x+x^2+x^3+x^4+x^5)(x+x^2+x^3+x^4+x^5)(x+x^2+x^3+x^4+x^5)(x+x^2+x^3+x^4+x^5)$

But why?

• Each of the four factors $(x+x^2+x^3+x^4+x^5)$ of the above product correspond to a room.
• The term $x^i$ in each factor $(x+x^2+x^3+x^4+x^5)$ correspond to allocating the respective room $10\times i$ chairs.
• Finally we require that the sum of chairs in all the rooms is $100$ , which correspond to a $x^{10}$ term in the final expression.

Note: The four factors are multiplied in the above expression so that when we multiply different terms of the form $x^i$ , their powers add. ( End of Note )

This means that the number of ways to distribute the chairs is equal to the number of $x^{10}$ terms in the final expression, which is same as the coefficient of $x^{10}$ in the product $(x+x^2+x^3+x^4+x^5)^4$ . In the calculation that follows, read $[x^i]p(x)$ as coefficient of $x^i$ in $p(x)$ .

$[x^{10}](x+x^2+x^3+x^4+x^5)^4$

$=[x^{10}]x^4(1+x+x^2+x^3+x^4)^4$

$=[x^{6}](1+x+x^2+x^3+x^4)^4$

$=[x^{6}](1-x^5)^4(1-x)^{-4}$

$=[x^{6}](1-x^5)^4(1-x)^{-4}$

$=[x^{6}](1-4x^5+6x^{10}-4x^{15}+x^{20})(1-x)^{-4}$

as the terms $6x^{10} \text{, }-4x^{15}\text{ and }x^{20}$ won't contribute in creation of $x^6$ term, the expression becomes-

$[x^{6}](1-4x^5)(1-x)^{-4}$

$=[x^{6}](1-x)^{-4} - 4\times [x^{6-5}](1-x)^{-4}$

as $[x^i](1-x)^{-r} = \binom{i+r-1}{r-1}$ , we can write the above as-

$\binom{6+4-1}{4-1} - 4\binom{1+4-1}{4-1} = \binom{9}{3} - 4\binom{4}{3} = 84-16 = 68$