# A school debating team consisting of exactly four students is to be selected from amongst twelve eligible students. Jane and Mike are among the twelve eligible students.Find the number of ways in...

A school debating team consisting of exactly four students is to be selected from amongst twelve eligible students. Jane and Mike are among the twelve eligible students.Find the number of ways in which the team can be selected

(2) Either Jane or Mike in the team

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A 4 person team is to be selected from 12 students with Jane and Mike among the 12.

How many teams have either Jane or Mike on them?

Usually we take "or" to be the inclusive or -- either Jane or Mike or both.

The number of teams with Jane on them (Mike not) is `_10C_3=120`

The number of teams with Mike on them (Jane not) is also 120.

The number of teams with Jane and Mike on them is `_10C_2=45`

The number of teams with Jane, Mike, or both is found by taking the sum so there are 240+240+45=285 teams.

Alternatively, there are `_11C_3=165` teams with Jane on them and 165 teams with Mike on them.

To find the number with Jane, Mike, or both we must subtract the overlap (those teams with Jane that also have Mike and vice versa). The number of overlaps is 45. So the total teams with Jane, Mike, or both is 165+165-45=285 as before.

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**Assuming that the "or" is inclusive there are 285 teams with Jane, Mile, or both on them.**

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If the "or" is meant to be exclusive, then the answer is 240 -- we found the number with just Jane or Mike to be 120 each so the total is 240.

Here we have to choose either Jane or Mike but not both. So we have to include one of them in the group and omit the other and select another 3 from the remaining 10.

Groups with Jane included `= ^10C_3 = 120`

Groups with Mike included `= ^10C_3 = 120`

*so there will be 240 groups with either Jane or Mike included in the team.*