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

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

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15 Sep 2017, 02:04
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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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15 Sep 2017, 05:04
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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15 Sep 2017, 12:20
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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15 Sep 2017, 12:26
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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15 Sep 2017, 13:57
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

**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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17 Sep 2017, 18:46
=>

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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20 Sep 2017, 15:53
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.

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Re: If 3 persons are selected at random from 8 persons for 3 positions pr   [#permalink] 20 Sep 2017, 15:53
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