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ketandoshi: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sat May 15, 2010 3:30 pm; Location: Houston, TX; Thanked: 1 times

Question from Kaplan Premier 2010-2011

by ketandoshi » Thu Aug 12, 2010 7:58 pm

Can somebody help me to understand this problem.

A company has 13 employees, 8 of whom belong o the union. If 5 people work 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?

Thanks in advance.

Ketan

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Source: Beat The GMAT — Problem Solving | Thu Aug 12, 2010 7:58 pm

sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

by sanju09 » Fri Aug 13, 2010 12:09 am

ketandoshi wrote:Can somebody help me to understand this problem.

A company has 13 employees, 8 of whom belong o the union. If 5 people work 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?

Thanks in advance.

Ketan

If at least 4 union members have to work each shift, then the many different combinations of employees might work any given shift

= 8C5 Ã— 5C0 + 8C4 Ã— 5C1

= 56 Ã— 1 + 70 Ã— 5

= 56 + 350

= 406

Last edited by sanju09 on Fri Aug 13, 2010 1:37 am, edited 1 time in total.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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limestone: Master | Next Rank: 500 Posts; Posts: 307; Joined: Sun Jul 11, 2010 7:52 pm; Thanked: 36 times; Followed by:1 members; GMAT Score:640

by limestone » Fri Aug 13, 2010 1:08 am

Sanju09, I'm a little confused here. In your answer, you splitted the problem into 2 cases.

The first: all the people in a shift are from the union - 8C5 x 5C0 = 56 x1 = 56

The second: only four in the shift are from the union, one is from the out-of-the-union employees - 8C4 x 5C1 = 70 x 5 = 350
( 8C4 = 70, am I right?)

The total different combinations: 406

However, I try to solve it in another way, and find out that the result does not match. Could anyone pls check if there is any error here.

There must always be at least 4 from the union in a given shift. So I pick out 4 of 8 that belong to the union - 8C4 = 70

I add the remain 4 to 5 that does not belong to the union, then choose 1 from that sum to combine with the previously choosen 4. 9C1 = 9

The total combinations here is: 70 x 9 = 630 - Oh oh, something wrong here?

I can not find out any thing wrong in both method.[/quote]

Quote

sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

by sanju09 » Fri Aug 13, 2010 1:29 am

limestone wrote:Sanju09, I'm a little confused here. In your answer, you splitted the problem into 2 cases.

The first: all the people in a shift are from the union - 8C5 x 5C0 = 56 x1 = 56

The second: only four in the shift are from the union, one is from the out-of-the-union employees - 8C4 x 5C1 = 70 x 5 = 350
( 8C4 = 70, am I right?)

The total different combinations: 406

However, I try to solve it in another way, and find out that the result does not match. Could anyone pls check if there is any error here.

There must always be at least 4 from the union in a given shift. So I pick out 4 of 8 that belong to the union - 8C4 = 70

I add the remain 4 to 5 that does not belong to the union, then choose 1 from that sum to combine with the previously choosen 4. 9C1 = 9

The total combinations here is: 70 x 9 = 630 - Oh oh, something wrong here?

I can not find out any thing wrong in both method.

Yes, you are right. 8C4 = 70, my mistake!

But how the remaining 4 from 8 do not belong to the union when they in fact do? Your work in red is fallacious and the correct answer is 406.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

Quote

ketandoshi: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sat May 15, 2010 3:30 pm; Location: Houston, TX; Thanked: 1 times

by ketandoshi » Fri Aug 13, 2010 6:52 am

Thanks Sanju. I like the way you split the problem.

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selfmade: Senior | Next Rank: 100 Posts; Posts: 82; Joined: Thu May 27, 2010 12:23 pm; Location: Newyork; Thanked: 2 times; Followed by:3 members

by selfmade » Fri Aug 13, 2010 9:04 am

This is a good example.

----------
Aiming for 780

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soham2208: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Tue Aug 10, 2010 9:53 am; Location: MUMBAI India; Thanked: 1 times; GMAT Score:750

by soham2208 » Tue Aug 17, 2010 7:17 am

limestone wrote:Sanju09, I'm a little confused here. In your answer, you splitted the problem into 2 cases.

The first: all the people in a shift are from the union - 8C5 x 5C0 = 56 x1 = 56

The second: only four in the shift are from the union, one is from the out-of-the-union employees - 8C4 x 5C1 = 70 x 5 = 350
( 8C4 = 70, am I right?)

The total different combinations: 406

However, I try to solve it in another way, and find out that the result does not match. Could anyone pls check if there is any error here.

There must always be at least 4 from the union in a given shift. So I pick out 4 of 8 that belong to the union - 8C4 = 70

I add the remain 4 to 5 that does not belong to the union, then choose 1 from that sum to combine with the previously choosen 4. 9C1 = 9

The total combinations here is: 70 x 9 = 630 - Oh oh, something wrong here?

I can not find out any thing wrong in both method.

[/quote]

When you use the second method - You are over-counting the number of cases. ! Here is an illustration of how -

Suppose the 8 members in union are - A B C D E F G H

Now, by 8C4 , suppose we get - A B C D, Then out of the remaining 9 people I select F...... I get the whole selection as A B C D F

Now consider another selection - A B C F, then out of the remaining 9 people I select D ...... I get the whole selection as A B C D F ... which is same as before... so similarly all such cases are re-counted....

Hence this approach is wrong....