Problem on combination - disarming a bomb with 10 wires I've been struggling with this combination problem for quite a while: 
A bomb has 10 wires. To defuse the bomb, 6 wires must be cut. However, among the 10 wires, there are 2 that if cut together the bomb detonates. Of how many ways can you cut the wires without detonating the bomb? 
It says that the answer is 140, but I can't get to this value. Could anyone clarify what I'm not getting here?
 A: It's easier to calculate the number of ways that detonate than to calculate directly those that don't.
The total number of ways to choose $6$ wires of the $10$ to cut is ${10 \choose 6}$.  Of these, the number of choices that detonate the bomb, i.e. those that consist of the $2$ special wires plus $4$ of the other $8$, is ${8 \choose 4}$.
The difference is ${10 \choose 6} - {8 \choose 4}$, which is indeed $140$.
A: Here is a direct calculation:  To disarm the bomb, we must either cut six of the eight wires that cannot set off a detonation or exactly one of the two wires that cause the bomb to detonate when both are cut and five of the other eight wires.
$$\binom{8}{6} + \binom{2}{1}\binom{8}{5} = 140$$
A: Let $w_1,\ldots,w_{10}$ be the 10 wires, and without loss of generality assume the two wires that should both not be cut are $w_1$ and $w_2$.  The number of ways to cut 6 wires without cutting both $w_1$ and $w_2$ is equal to the sum of three quantities: (a) the number of ways to cut 6 wires so that $w_1$ is cut and $w_2$ is not cut (this can be done in ${8 \choose 5}$ ways because the remaining 5 wires must be chosen from $\{w_3,\ldots,w_{10}\}$), (b) the number of ways to cut 6 wires so that $w_2$ is cut and $w_1$ is not cut (this quantity is also ${8 \choose 5}$, and (c) the number of ways to cut 6 wires so that neither $w_1$ nor $w_2$ is cut (this is ${8 \choose 6}$.  The sum of these three values is $140$.
