I have a question in the lecture that was answered by the lecturer. But then again, it was just briefly explained and i still cannot digest the idea

The question is count the subsets of the set X = {A,B,C,D,E,F,G} whose intersection with {A,B,C,D} has size 2 and the answer is 4C2 x 8 = 48. I was thinking about it but do not really understand why the first term relates to the second and it makes me so confused and wonder that “what am i counting now”

Also, do you have any general method to solve this kind of intersection of subsets ?

Thanks so much for your help in advance

  • 1
    $\begingroup$ You're choosing which $2$ of the $4$ candidate elements ($A,B,C,D$) are in the intersection (${{4}\choose{2}}=6$) and then which of the remaining $3$ elements of $X$ ($E,F,G$) are also in the subset ($2^3=8$). $\endgroup$ – mjqxxxx Apr 8 '18 at 15:36

You can construct each possible set $S$ that satisfies the requirements by first choosing $Y = S \cap \{A, B, C, D\}$ and then choosing $Z = S \cap \{E, F, G\}$. Distinct choices of $Y$ or $Z$ lead to distinct $S$. There are $_4C_2 = 6$ choices for $Y$ and $2^3 = 8$ choices for $Z$ leading to $6 \times 8 = 48$ possibilities for $S$.

  • $\begingroup$ Isnt it i have to find the subsets of X that satisfy the requirement? What do you mean by “forming X”?. Why we have to do that? that’s also the point that leads me to be foolish in this question :( . Thank you $\endgroup$ – DinhHoang Apr 8 '18 at 16:01
  • $\begingroup$ I'm describing all the ways to construct (or form) the set $X$. $\endgroup$ – Rob Arthan Apr 8 '18 at 16:32
  • $\begingroup$ @DinhHoang: I've tried to clarify the answer. Does it make sense now? A very common way of addressing this kind of combinatorial problem is to describe a process that constructs all the objects you are trying to count. $\endgroup$ – Rob Arthan Apr 8 '18 at 17:31
  • $\begingroup$ It is clear now. Thank you so much!! $\endgroup$ – DinhHoang Apr 9 '18 at 2:39

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