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Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

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01 Sep 2015, 23:23

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Bunuel wrote:

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected............?

A) 28 B) 46 C) 55 D) 63 E) 70

Ans:C

Solution: J+T+6 = total 8 , we need to select at least one of the J and T or both= total 4 out of 8 - 4 out of 6 8C4 - 6C4 =55
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No of ways to form committee with only Jane (not Thomas) : J _ _ _ ---> 6C3 = 20 No of ways to form committee with only Thomas (not Jane) : T _ _ _ ---> 6C3 = 20 No of ways to form committee with both Thomas and Jane : T J _ _ ---> 6C2 = 15

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Bunuel wrote:

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28 B) 46 C) 55 D) 63 E) 70

Kudos for a correct solution.

Method 1: Number of committees in which Jane is selected is 7C3 or 35 Number of committees in which Thomas is selected is 7C3 or 35 Number of committees in which both are selected is 6C2 or 15 Number of committees in which at least one of either Jane or Thomas is selected will be

(A u B) = A + B - ( A n B) = 35 + 35 -15 = 55

Method 2: Total number of committees that can be formed is 8C4 or 70 Total number of committees that can be formed which do not include Jane and Thomas willl be 6C4 or 15 So, the number of committees in which at least one of either Jane or Thomas is selected will be

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02 Sep 2015, 17:42

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Bunuel wrote:

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28 B) 46 C) 55 D) 63 E) 70

Kudos for a correct solution.

Number of ways of selecting 'r' from total 'n' is given by formula ncr = n!/(r!(n-r)!) Oder is not important in forming a committee. Number of ways if at least one of Jane or Thomas is selected = Number of ways (Jane only is selected + Thomas only is selected + both Jane and Thomas are selected)

So, if Jane is selected in the committee, number of ways other 3 members are selected from group of 6 (8 - Jane and Thomas) (J _ _ _)= 6C3 = 6!/(3!*(6-3)!) = 20 ways if Thomas is selected in the committee, number of ways other 3 members are selected from group of 6 (8 - Jane and Thomas) (T _ _ _) = 6C3 = 6!/(3!*(6-3)!) = 20 ways if both Jane and Thomas are selected in the committee, number of ways other 2 members are selected from group of 6 (8 - Jane and Thomas) (T J _ _) = 6C2 = 6!/(2!*(6-2)!) = 15 ways

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

The first step in this problem is to recognize that order does not matter. We must create a committee of four people, but we are not putting these people in any sort of order. Because order does not matter, we must use the combinations formula, which is n!/[k!(n-k)!], in order to determine the number of possible outcomes.

However, as is the case with most advanced combinations problems, the solution is not as simple as plugging a few numbers into the formula. Instead, we need to consider exactly what types of outcomes we are looking for.

In this case, we need to select Jane and 3 other people, none of whom are Thomas; Thomas and 3 other people, none of whom are Jane; or Jane, Thomas and 2 other people. Additionally, we need to note that this is an ‘or’ situation, which means that we need to add all of these possibilities together.

First, if we select Jane and three others, none of whom are Thomas, we must pick Jane plus 3 out of the remaining 6 people. Because we can pick any 3 out of the 6, we set n = 6 and k = 3. When we plug this into the formula it gives us 6!/(3!3!) = 20 (Quick review: 3! = 3 x 2 x 1, 6! = 6 x 5 x 4 x 3 x 2 x 1.)

Next we determine how many ways we can select three people, none of whom are Jane, to go along with Thomas. This is identical mathematically, thus we end up with 20 in this case too.

Lastly, we determine how many outcomes are possible in which we have Jane, Thomas and 2 others. We now select 2 out of 6, so n = 6 and k = 2. 6!/(4!2!) = 15.

All we need to do to get the answer is add up all of the possible outcomes. 20 + 20 + 15 = 55, which is the correct answer to this question (C).
_________________

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28 B) 46 C) 55 D) 63 E) 70

If at least one of either Jane or Thomas is selected for the committee, that means that either Jane is selected, Thomas is selected, or both are selected.

If Jane but not Thomas is selected, there are 6 people remaining for 3 spots. Thus, when Jane is selected, the committee can be chosen in 6C3 = (6 x 5 x 4)/3! = 20 ways. Similarly, if Thomas but not Jane is selected, the committee can be chosen in 20 ways.

Finally, if they are both selected, then there are 6 people remaining for 2 spots. Thus, the committee can be chosen in 6C2 = (6 x 5)/2! = 15 ways.

Therefore, the number of ways the committee can be chosen with at least one of either Jane or Thomas is 20 + 20 + 15 = 55 ways.

Answer: C
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