The following two questions seem similar to me, but according to the solutions, they are both solved in a different way.

Shouldn't they be solved in the same way? If so, which is the correct method?

Question 1:

How many 4-person committees are possible from a group of 9 people if, either John or Barbara (but not both) must be on the committee?



1C1 * 7C3 + 1C1 *7C3 = 70

Question 2:

A basketball team has 5 distinct positions. Out of 8 players, how many starting teams are possible if, the distinct positions are not taken into consideration, but either Mike or Ken (but not both) must start?


7C4 + 7C4 - 6C3 = 50

7C4: Number of starting combinations involving Mike or Ken

6C3: Number of starting combinations involving both Mike and Ken


I think both of your solutions are wrong.

Answer 1 -

We have to pick 1 from John or Barbara.

$$\binom 21$$

And remaining 3 from remaining 7

$$\binom 73$$

So we have,

$$\binom 21 \times \binom 73$$

Similarly for second part.

  • $\begingroup$ Yes, the solution for the 1st one was wrong (I had typed the answer for another question) - it has been updated. Why is the second one wrong? $\endgroup$ – kpatelio Apr 2 '17 at 18:55
  • 1
    $\begingroup$ @kpatelio You added all the teams that include Ken and all the teams that include Mike, which means you counted all the teams that include both Ken and Mike twice. You subtracted these teams once, which means that you are still counting teams that include both Ken and Mike once. However, you do not want to count them at all. Therefore, the number of teams that include neither Ken or Mike is $$\binom{7}{4} + \binom{7}{4} - 2\binom{6}{3} = 30$$ If you use the method outlined by Kanwaljit Singh, you will obtain $$\binom{2}{1}\binom{6}{4} = 30$$ $\endgroup$ – N. F. Taussig Apr 2 '17 at 19:11

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