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Expert replies
by adssaini » Wed Aug 05, 2009 11:59 am
a. A five member committee is to be selected from among four Math teachers and five English teachers. In how many different ways can the committee be formed under the following circumstance?

1)The committee must contain at least three Math teachers.

2) The committee must contain at least three English teachers.

sol : 1) 45 2)81



b. From a group of 8 teachers, a committee of at least one and at most three persons is to be formed. How many different committees can be formed? (sol . 92)

c.
If there are 8 orange bars, 9 red bars and 5 blue bars, how many different ways are there to give a person 2 orange bars, 3 red bars and 1 blue bar? (sol. 11760)


Any body please explain how to solve these....
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Source: — Problem Solving |

by gmat740 » Wed Aug 05, 2009 12:24 pm
I would go one by one.

a.)
1. 4M and 5E
we have to select a committee of 5
so, we can have either 3M or 4M

thus,
4C3*5C2 + 4C4*5C1 = 40 +5 = 45

2.) We can have 3E,4E or 5E

5C3*4C2 + 5C4*4C1 + 5C5*4C0 = 60 +20 +1 = 81
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by gmat740 » Wed Aug 05, 2009 12:29 pm
b. From a group of 8 teachers, a committee of at least one and at most three persons is to be formed. How many different committees can be formed? [spoiler](sol . 92)[/spoiler]
There can be one or two or 3 teachers in a committee :

8C1 + 8C2 + 8C3 = 8 +28 +56 = 92

Hope this helps
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by gmat740 » Wed Aug 05, 2009 12:38 pm
c.
If there are 8 orange bars, 9 red bars and 5 blue bars, how many different ways are there to give a person 2 orange bars, 3 red bars and 1 blue bar? [spoiler](sol. 11760)[/spoiler]
PLEASE USE SPOILERS

8C2*9C3*5C1 = 11760

Simple questions

You need to go through the basics of P&C.

Hope this helps
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