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There are 26 members. We want to make a team of 4 members from that. Let, a and b be two members from that 26 members.

i) How many different are teams possible having a and b,

ii) How many different teams are possible without having a and b?

I think the result of (i) should be $ C (24,2) $ And result of (ii) should be $ C (24, 4) $

Shouldn't they?

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    Consider a team that has $a$ but not $b$. – 2017-01-25

3 Answers 3

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For I, you take two people from the the group not containing A en B. I assume that order of the teams does not matter here, otherwise you also have to take this into mind. This implies that we have $1\cdot1\cdot24\cdot23=552$ teams.

For II, you have to take 4 people from the group not containing a and b, so we have to choose 4 people from a group of 24 people. How can we do this?

By also solving II, you can easily see that the third exercise is not true.

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    Can you mathematically explain why have you written $ 1 . 1 . 24 . 23 teams$ as an answer to ques i – 2017-01-25
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    Considering the order does not matter, two of the four people must be A and B. For the other two, we have to choose from the group of 24 people. For the first one we have 24 choices. After choosing we are left with 23. So the totel combination is $24\cdot23$. – 2017-01-25
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For i, you just pick two other people to fill out the committee from the remaining 24. For ii, you pick four from the remaining 24. Clearly iii is not true. Can you see why?

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    yes obviously it is false, when I mathematically calculate the answers of (i) and (ii) to be $C (24,2)$ and $ C (24,4)$. But, logically (iii) appears to be true. Because, out of total possible teams, if some of them contains a and b, and rest doesn't contain a and b. – 2017-01-25
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    Oh extremely sorry! I got my mistake watching @Joffan 's comment – 2017-01-25
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i) number of teams = 24c2 = 276

ii) number of teams = 24c4 = 10626

iii) NO, you can make a team with only a and not b, or you can make a team with only b

actually you can make 26c4 = 14950 teams which doesn't equal i+ii

c is combinations https://en.wikipedia.org/wiki/Combination