Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I am just stuck with a problem and need your help. The question is:
A closet has 5 pairs of shoes. The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair is a.80 b.160 c.200 d.none
I think the answer should be $\binom{10}{1}\cdot\binom{8}{1}\cdot\binom{6}{1}\cdot\binom{4}{1}$. But I seriously doubt this could be the answer. Please help me in solving this.
Hint: The four shoes must come from 4 different pairs.
For a specific set of 4 pairs, there are $2^4$ ways of choosing the shoes. There are 5 possible sets of 4 pairs. Hence there are $5*2^4$ ways in total.
Actually your idea is right, but you are counting the number of ordered sequences of four shoes. Divide by $4!$ and you get the answer you are looking for.
We need to chose $4$ different color shoes from the five different pairs of shoes given. Therefore number of ways to select them is $\displaystyle \binom{5}{4}$. Now from a pair of two shoes we need to select only one shoe so that we don't get a complete pair of shoes in our four selected $4$ shoes. The number of ways to select them is $\displaystyle \binom{2}{1}$. We need to do this with $4$ different times Therefore their number of ways becomes $\displaystyle\binom{2}{1}^4$. Therefore the total number of ways to choose $4$ shoes will be $\displaystyle \binom{5}{4}\cdot\binom{2}{1}^4$ which is equal to $80$.
Since we have $5$ pairs of shoes, and we wish to choose $4$ shoes avoiding pairs of shoes, the answer is the number of independent $4$-sets of vertices of the graph:
I.e., it is the coefficient of $x^4$ in its independence polynomial.
We know that the independence polynomial of a disconnected graph is equal to the product of independence polynomials of its connected components. The independence polynomial of $K_2$ is $1+2x$, so the independence polynomial of the above graph is $$(1+2x)^5=32x^5+80x^4+80x^3+40x^2+10x+1.$$ Hence there are $80$ independent sets of size $4$.
