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!
$\endgroup$ 52 Answers
$\begingroup$number of subsets in which $a$ must present (and $b$ may be or mayn't be present) is $ \displaystyle {9\choose 3}=84$
similarly, number of 4 element subsets in which $b$ must present (and $a$ may be or mayn't be present) is $ \displaystyle {9\choose 3}=84$
and 4 element subsets which contains $a$ as well as $b $ is $\ \displaystyle {8\choose 2}=28$
number of subsets which contains $a$ OR $b$ is $= (2\times 84)-28=140$
we know,
$\text {atleast} = \text{total} - \text{none}$
therefore number of $4$ element subsets contain atleast one of $a$ and $b$ $= 210- \displaystyle {8\choose 4} =140$
$\endgroup$ $\begingroup$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.
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