How many $4$-element subsets of $S = \{a, b, c, d, e, f, g, h, i\}$ contain at least one of $a$ and $b$? Consider the set $S = \{a,b,c,d,e,f,g,h,i,j\}$.  How many $4$-element subsets of $S$ contain at least one of $a$ and $b$?
So my thought is :
1) number of $4$-element subsets in the set $S = 210$
2) $4$-element subsets of $S$ contain exactly one of $a$ and $b$ $= \dfrac{2!}{1!1!} \cdot \dfrac{8!}{3!5!} = 112$
3) $4$-element subsets of $S$ contain both $a$ and $b$ $= C(8,2) = \dfrac{8 \cdot 7}{2 \cdot 1} = 28$
How many $4$-element subsets of $S$ contain at least one of $a$ and $b$?
\begin{align*}
& = 210 - (112+28)\\
& = 70
\end{align*}
Is it right?
thanks!

You have correctly calculated that the number of four-element subsets of the ten-element set $S$ is 
$$\binom{10}{4} = 210$$
that the number of subsets of $S$ that contain exactly one of $a$ or $b$ is
$$\binom{2}{1}\binom{8}{3} = 112$$
and that the number of subsets of $S$ that contain both $a$ and $b$ is 
$$\binom{2}{2}\binom{8}{2} = 28$$
However, you have drawn the wrong conclusion.  The number of subsets that contain at least one of $a$ and $b$ is the sum of the number that contain exactly one of $a$ or $b$ and the number that contain both $a$ and $b$.  Hence, the number of subsets that contain at least one of $a$ and $b$ is 
$$\binom{2}{1}\binom{8}{3} + \binom{2}{2}\binom{8}{2} = 140$$
Alternatively, we could subtract the number of subsets of $S$ that contain neither $a$ nor $b$ from the total.  The number of subsets of $S$ that contain neither $a$ nor $b$ is 
$$\binom{2}{0}\binom{8}{4} = 70$$
Subtracting this from the number of four-element subsets of $S$ yields
$$\binom{10}{4} - \binom{2}{0}\binom{8}{4} = 140$$
as before.
