Question from Kaplan Premier 2010-2011

[ketandoshi]

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

[sanju09]

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 × 5C0 + 8C4 × 5C1

= 56 × 1 + 70 × 5

= 56 + 350

= 406

[limestone]

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]

[sanju09]

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.

[ketandoshi]

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

[selfmade]

This is a good example.

[soham2208]

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....

[HPengineer]

disregard my posti got it now..