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Company A has 13 employees, 8 of whom belong to the union. If 5 people

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u0422811
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Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   Updated on: 08 Jul 2015, 01:55
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Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?

(A) 56
(B) 231
(C) 336
(D) 350
(E) 406
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E

Originally posted by u0422811 on 29 Sep 2010, 21:30.
Last edited by Bunuel on 08 Jul 2015, 01:55, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   29 Sep 2010, 21:48
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u0422811 wrote:
Company A is twice the size of its nearest competitor, Company B. Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?

(a) 56
(b) 231
(c) 336
(d) 350
(e) 406


Looks easy: -- 8C4*5C1 + 8C5 -- 406 (E).
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Re: Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   30 Sep 2010, 02:23
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aarthis2 wrote:
ezhilkumarank wrote:
u0422811 wrote:
Company A is twice the size of its nearest competitor, Company B. Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?

(a) 56
(b) 231
(c) 336
(d) 350
(e) 406


Looks easy: -- 8C4*5C1 + 8C5 -- 406 (E).


Could you please explain how you got 8c5.I used this logic...Totally 13 employees out of which 8 are union emp.In order to have atleast 4union employees in any shift.I splitted it as 8c4*5c1(selecting 4 emp from 8 union emp and 1 employee from remaining 5 non-union employees).I got the answer as 350.I didnot understand how you got 8c5.Kindly help me.


There are total of 13 employees, 8 of whom belong to the union and 5 doesn't. Out of 5 people working a shift at least 4 must belong to the union.

Now, at lest 4 out of 5 means that 4 or all 5 employees must belong to the union:
Out of 5 people working a shift 4 employees belong to the union and 1 doesn't: \(C^4_8*C^1_5=350\);
Out of 5 people working a shift all 5 employees belong to the union: \(C^5_8=56\);

Total # of ways: \(350+56=406\).

Answer: E.

Hope it's clear.
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Re: Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   07 Jul 2015, 22:52
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u0422811 wrote:
Company A is twice the size of its nearest competitor, Company B. Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?

(a) 56
(b) 231
(c) 336
(d) 350
(e) 406


Hi Bunuel,

The opening sentence to this prompt has absolutely no bearing on the actual question that is asked. I have to assume that it was originally included by accident - since the Official GMAT would not include this type of 'filler', can you remove that initial sentence?

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Re: Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   30 Sep 2010, 02:08
ezhilkumarank wrote:
u0422811 wrote:
Company A is twice the size of its nearest competitor, Company B. Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?

(a) 56
(b) 231
(c) 336
(d) 350
(e) 406


Looks easy: -- 8C4*5C1 + 8C5 -- 406 (E).


Could you please explain how you got 8c5.I used this logic...Totally 13 employees out of which 8 are union emp.In order to have atleast 4union employees in any shift.I splitted it as 8c4*5c1(selecting 4 emp from 8 union emp and 1 employee from remaining 5 non-union employees).I got the answer as 350.I didnot understand how you got 8c5.Kindly help me.
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Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   20 Mar 2016, 08:06
Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?

Total combinations= 4 from 8 union * 1 from 5 rest + 5 from 8 union
=8C4*5C1 + 8C5
= 350+56
=406
Hence E
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Re: Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink] New post   24 Dec 2023, 01:07
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Re: Company A has 13 employees, 8 of whom belong to the union. If 5 people [#permalink]
24 Dec 2023, 01:07
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