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If 1 male, 2 females, and 1 child are to be randomly selected from 8 m

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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
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If 1 male, 2 females, and 1 child are to be randomly selected from 8 m  [#permalink]

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New post 13 Feb 2017, 00:00
1
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

83% (01:29) correct 17% (02:13) wrong based on 38 sessions

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If 1 male, 2 females, and 1 child are to be randomly selected from 8 males, 10 females, and 8 children, how many such cases are possible?

A. 980
B. 1,440
C. 1,880
D. 2,480
E. 2,880

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Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m  [#permalink]

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New post 13 Feb 2017, 00:38
MathRevolution wrote:
If 1 male, 2 females, and 1 child are to be randomly selected from 8 males, 10 females, and 8 children, how many such cases are possible?

A. 980
B. 1,440
C. 1,880
D. 2,480
E. 2,880

Hi,

Few Important points:

AND => Multiplication

OR => Addition.

When Order is not important => Apply Combination formula

When Order is important => Apply Permutation formula.

If 1 male, 2 females, and 1 child are to be randomly selected from 8 males, 10 females, and 8 children, how many such cases are possible?

In the above problem order is not important, so we'll use Combination formula.

We have to select

1 male from 8 males, AND 2 females from 10 females, AND 1 child from 8 children.

= \(^{8}C_1 \times ^{10}C_{2} \times ^{8}C_{1} = 8 \times 45 \times 8 = 2880\)

Thanks.
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Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m  [#permalink]

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New post 13 Feb 2017, 01:16
1 male can be chosen from 8 males in 8 ways;

1 child can be chosen from 8 children in 8 ways;

2 females can chosen from 10 females:

1. 1st female can be chosen in 10 ways; There are 9 females remaining;
2. so, 2nd female can be chosen in 9 ways;
Now the order does not matter;
so, total possible selection for 2 females (10*9)/2! = 45;

Therefore, total possible cases = 8*8*45 = 2880; Answer E.
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Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m  [#permalink]

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New post 15 Feb 2017, 00:08
==> You get 8C1*10C2*8C1=(8)(45)(8)=2,880.

The answer is E.
Answer: E
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Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m &nbs [#permalink] 15 Feb 2017, 00:08
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If 1 male, 2 females, and 1 child are to be randomly selected from 8 m

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