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# The three competitors on a race have to be randomly chosen from a grou

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Math Expert
Joined: 02 Sep 2009
Posts: 49988
The three competitors on a race have to be randomly chosen from a grou  [#permalink]

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16 Oct 2016, 04:47
00:00

Difficulty:

35% (medium)

Question Stats:

69% (01:48) correct 31% (01:40) wrong based on 118 sessions

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The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios contain at least one woman?

A. 10
B. 15
C. 16
D. 30
E. 46

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Re: The three competitors on a race have to be randomly chosen from a grou  [#permalink]

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16 Oct 2016, 05:41
Bunuel wrote:
The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios contain at least one woman?

A. 10
B. 15
C. 16
D. 30
E. 46

Total ways of choosing the runners randomly = (5+3)C3 = 8C3 = 56

Unwanted cases = team of 3 MALE runners = 5C3 = 10

Desired cases = 56 - 10 = 46

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Re: The three competitors on a race have to be randomly chosen from a grou  [#permalink]

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18 Nov 2017, 16:56
1
Hi All,

We're told to choose three competitors from a group of five men and three women. We're asked for the number of different trios that include AT LEAST one woman. The answer choices to this question are 'spaced out' enough that you can do a little bit of work to eliminate all of the wrong answers.

To start, if the group has just 1 woman, then there will be 2 men. The number of groups of 2 men is 5c2 = 5!/(2!)(3!) = 10 different pairs of men.

With 3 women to choose from, there would be 3(10) = 30 groups with just 1 woman. There would then be additional groups with 2 women or all 3 women, so the total number of groups MUST be greater than 30. There's only one answer that's possible...

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Re: The three competitors on a race have to be randomly chosen from a grou  [#permalink]

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22 Nov 2017, 13:25
2
Bunuel wrote:
The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios contain at least one woman?

A. 10
B. 15
C. 16
D. 30
E. 46

We can use the formula:

Total number of ways to select the group - number of ways with no women = number of ways with at least one woman.

Total number of ways to select 3 people from a group of 8 people:

8C3 = 8!/[3!(8-3)!] = 8!/3!5! = (8 x 7 x 6)/3! = 56

Number of ways with no women:

5C3 = 5!/[3!(5-3)!] = (5 x 4 x 3)/3! = (5 x 4 x 3)/(3 x 2 x 1) = 10

Thus, the number of ways with at least one woman is 56 - 10 = 46.

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Re: The three competitors on a race have to be randomly chosen from a grou &nbs [#permalink] 22 Nov 2017, 13:25
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