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If 3 persons are selected at random from 8 persons for 3 positions pr

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If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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[GMAT math practice question]

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680
[Reveal] Spoiler: OA

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Re: If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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New post 15 Sep 2017, 04:04
MathRevolution wrote:
[GMAT math practice question]

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680


Choose for present position: 8 ways.
Choose for vice-present position: 7 ways
Choose for secretary position: 6 ways.

Total possible ways: 8 * 7 * 6 = 336 ways. Answer D
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Re: If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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New post 15 Sep 2017, 11:20
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No of possibilities when 3 people are selected out of 8 = 8C3 = 56
Now further, the positions can be arranged in 3! ways = 56*3! = 336 (D)
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Re: If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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New post 15 Sep 2017, 11:26
MathRevolution wrote:
[GMAT math practice question]

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680


Can be solved through Combination, however this is an arrangement question where 8 people need to be arranged in 3 positions.
Hence \(8_P_3\) \(=336\)

Option D
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If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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New post 15 Sep 2017, 12:57
MathRevolution wrote:
[GMAT math practice question]

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680

Order matters. President is like first prize, vice-president is like second prize, and secretary is like third prize.

It's a permutation (or use FCP because there are three slots, and 8*7*6 possibilities).

\(_{n}P_{k} = \frac{n!}{(n - k)!}\)

\(_8P_3 = \frac{8*7*6*(5!)}{(5!)}\) = 336

ANSWER D

**Long version

\(_8P_3 = \frac{8*7*6*5*4*3*2*1}{5*4*3*2*1}\) = 336
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Re: If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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New post 17 Sep 2017, 17:46
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8P3 = 8 x 7 x 6 = 336

Ans: D
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Re: If 3 persons are selected at random from 8 persons for 3 positions pr [#permalink]

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New post 20 Sep 2017, 14:53
MathRevolution wrote:
[GMAT math practice question]

If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?

A. 56
B. 70
C. 240
D. 336
E. 1680


The order of selection is important in this scenario, so this is a permutation problem. The number of ways to select 3 people from 8 for the given positions is 8P3 = 8!/(8-3)! = 8 x 7 x 6 = 336 ways.

Answer: D
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Re: If 3 persons are selected at random from 8 persons for 3 positions pr   [#permalink] 20 Sep 2017, 14:53
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