Question 477451: Dear math teacher,
I am having difficulties solving the following problem (#92b)in the fastest and least time consuming way. I got the result but when I tried the second method, my answers did not match. Would you please explain why?
Here is the problem:
"91. From 5 physicists, 4 chemists and 3 mathematicians a committee of 6 is to be chosen so as to include 3 physicists, 2 chemists and 1 mathematician. In how many ways can this be done?"
"92. In problem 91, in how many ways can the committee of 6 be chosen so that
a) 2 members of the committee are mathematicians,
b) at least 3 members of the committee are physicists?
P Chem Math
n 5 4 3
r 3
4
5
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3 3 3 thus, 5C3 times 4C3 times 3C0 = 10.4.1 = 40
3 0 3 5C3.4C0.3C3 = 10.1.1 = 10
3 1 2 5C3.4C1.3C2 = 10.4.3 = 120
3 2 1 5C3.4C2.3C1 = 10.6.3 = 180
Total = 350
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4 2 0 5C4.4C2.3C0 = 30
4 0 2 5C4.4C0.3C2 = 15
4 1 1 5C4.4C1.3C1 = 60
Total = 105
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5 1 0 5C5.4C1.3C0 = 1.4.1 = 4
5 0 1 5C5.4C0.3C1 = 1.1.3 = 3
Total = 7
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350 + 105 + 7 = 462
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Now method 2:
Total # of ways to select committee of 6 out of 12 =
12C6 = 924 ways
Non-physicists can be selected in 12-5C6 ways = 7!/(7-6)!6! = 7 ways
I am not really sure how to continue solving this using this method. How can you correct me? Please let me know.
Thank you very much.
Yours sincerely,
Ivanka
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
b) at least 3 members of the committee are physicists.
possible ways are :
1. 3 physicists and 3 from chemists and mathematicians
2. 4 physicists and 2 from chemists and mathematicians
3. 5 physicists and 1 from chemists and mathematicians
total number of ways = 5C3 * 7C3 + 5C4 * 7C2 + 5C5 * 7C1
if any doubt, you are welcome to contact...
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