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Combination question !!!

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Combination question !!!

by nasir » Sun Sep 19, 2010 7:33 pm
a company 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 employees =13
# of unions employees =8
# of non union employees = 13-8= 5

we can select 4 employees from union workers = 8C4 = 8!/(8-4)!*4!
and 1 from non-union worker = 5C1 = 5!/(5-1)!*1!

=> 8C4 * 5C1 = 350

but the Answer explanation says that i have to add 8C5

8C5= the number of ways to select 5 union workers

i.e { 8C4 * 5C1) + 8C5 = 406 answer

i don't understand why we have to add 8C5.

Thanks

i posted this question and got its explanation as why we have to add 8C5.
Now there is another question.

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?

i tried this question on the same method, but got it wrong.
is there any similarity between the "quoted" question and this "Bold" question ? if not whats the difference ?

thanks
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by tpr-becky » Sun Sep 19, 2010 7:58 pm
The problem you are having with both set ups are that you are only considering one option. In the union question it says you have to have at least 4 union members. And you did the formula correctly for choosing 4 union members and 1 non-union. But you also have the option to have 5 union members. This is where the equation for 5 choose five comes from. When you have two different options for something then you need to add the possible combinations for each option.

IN the professor problem it says we have to have at least 1 professor which means you have to consider the situations with 1 professor, 2 professors and 3 professors and add each of them to get the total number of combination.

so you would have (4!/(4-1)!1!)(6!/(6-2)!2!) + (4!/(4-2)!2!)(6!/(6-1)!1!) + (4!/4-3!)3!) to get your answer.
Last edited by tpr-becky on Sun Sep 19, 2010 8:22 pm, edited 2 times in total.
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by limestone » Sun Sep 19, 2010 8:04 pm
In the quoted, 5 are selected from 13 ( 8 belong to the union, 5 do not) to form a group that has at least 4 belong to the union.
There're two cases:
4 from the union + 1 not from the union (in which no. of combination is 8C4*5C1)
5 all from the union ( in which no. of combination is 8C5)

In the "bold", 3 are selected from 10 ( 4 are professors, 6 are students) to form a group that has at least 1 professor.
There're such cases:
1 professor + 2 students : 4C1*6C2 = 4*15 = 60
2 professors + 1 student : 4C2*6C1 = 6*6 = 36
3 professors : 4C3 = 4
Total ways: 60+36+4 =100
However, there's another easier way to solve this. Find all possible combinations of 3 people, then eliminate combinations that consist no professor. We'll get combinations that have at least one professor.
pick 3 out of 6 students: 6C3 = 20
pick 3 out of total 10 people: 10C3 = 120
Total combinations that have at least one professor: 120 - 20 =100
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by kaushiksin » Mon Sep 20, 2010 5:24 am
professors - 4
students - 6

Committee must consist of 3 people.
If the committee has to include at least 1 professor
how many ways can this committee be formed?

Number of ways in which a committee can be formed =

3 Professors + 2 Professors, 1 Student + 1 professor, 2 Students

= 4C3 + 4C2*6C1 +4C1*6C2

=4 + 60 +36
= 100

The answer to this question will be 100.
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by nasir » Mon Sep 20, 2010 8:33 am
However, there's another easier way to solve this. Find all possible combinations of 3 people, then eliminate combinations that consist no professor. We'll get combinations that have at least one professor.
pick 3 out of 6 students: 6C3 = 20
pick 3 out of total 10 people: 10C3 = 120
Total combinations that have at least one professor: 120 - 20 =100
how would you solve the " Union- employees " question on this method ?
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by limestone » Mon Sep 20, 2010 6:28 pm
There're ten persons, 4 professors and 6 students. We're asked to pick a group of three that has at least one professors.
then, however we pick, there will be only 4 cases that happen:
a. No professor, 3 students
b. 1 professor, 2 students
c. 2 professors, 1 student
d. 3 professors, no student
Note that the total combinations can be formed when we pick 3 out of 10,"10C3", is the sum of combinations for all a,b,c,d cases above.
Or a+b+c+d = 10C3
Now there're 2 approaches this case:
Direct approach : b+c+d = 4C1*6C2+4C2*6C1+4C3 = 60 + 36 + 4 = 100
Indirect approach: 10C3 - a = 10C3 - 6C3 = 120 - 20 = 100
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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