10 persons are sitting in a circle. In how many ways can you select any three of them such that no two of them are consecutive?
My attempt: To get the required number, we subtract the unfavourable cases from the total number of cases.
The total number of cases will be the number of ways to select three people out of 10 people – 10C3 = 120
Out of these cases, unfavourable cases are those in which
(1) two of our selected people sit together while the third is sitting apart.
First we select any one of those people in 10C1 ways. To select a person sitting next to this person, we have 2C1 ways. Now, there are only 6 ways to select the third person. Therefore, the total number of ways is $10*2*6=120$.
(2) All three of them are sitting together:
There are 10 ways to do that.
The total number of unfavourable cases = 10+ 120 = 130
My unfavourable cases are exceeding the total number of cases.
What am I missing?
Best Answer
$\underline{A\; direct\; method}$
I take it that it is a numbered circle.
Make $3$ blocks of chosen-unchosen, looking clockwise, $\boxed{\circ\bullet}$ so there are now $3$ blocks and $4$ "singles"
The blocks can be placed among the $7$ units in $\binom73 = 35$ ways,
but each unit is getting only $7$ starting points instead of $10$,
thus the final answer $= \dfrac{10}{7} \times 35 = 50$