3 problems on Permutation-Combination

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3 problems on Permutation-Combination

by Uri » Tue Feb 03, 2009 9:02 am
Hi, please find below three questions on permutation-combination. I have got these from an online document, but feel that some of the answers may be wrong. So, it would be great if you can check whether my answer matches yours and then state your answers in reply. Thanks in advance.

Problem 1: If three cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the three cards if at least two cards are a jack, queen or a king?
OA: 2925
My Answer: 2860

Problem 2: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?
OA: 11760
My Answer: 117 :shock:

Problem 3: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 if it must contain at least three Math teachers?
OA: 41
My Answer: 45
Source: — Problem Solving |

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by hardik.jadeja » Tue Feb 03, 2009 10:04 am
Problem 2:
8C2*9C3*5C1 = 11760

I think you calculated three scenarios separately and then added them to get your final answer. But actually here we have to find how many possible combination are there for someone to get 6 bars (2 orange, 3 red and 1 blue). So it will be multiplication not addition..

Problem 3: even I am getting 45 here..

We have 4 math teachers and 5 English teachers..
We have to choose a committee of 5 such that there are at least 3 math teachers on that committee..

it can be done in 2 diff ways..
1) 3 Math Teachers and 2 English Teachers = 4C3*5C2 = 4*10 = 40
2) 4 Math Teachers and 1 English Teacher = 4C4*5C1 = 1*5 = 5

Total 40 + 5 =45

Can someone please check where am i going wrong if 45 is not the correct answer?

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by dendude » Tue Feb 03, 2009 1:40 pm
I agree with Hardik on P2 & P3.

For P1, I too get 2860,
(12C2*40C1)+(12C3) = 2860

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by DeepakR » Tue Feb 03, 2009 4:23 pm
Agreed with above. For P1 I am getting as follows:

If three cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the three cards if at least two cards are a jack, queen or a king?

Total Jack, Queen, King in 52 cards = 12.

At least two cards are a jack, queen or a king means = Either 2 cards are from J, K, Q and 1 card is from remaining cards OR (+) all 3 cards are from J, Q, K.

Hence Total= 12C2 * 40C1 + 12C3 = 2860.

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by Uri » Wed Feb 04, 2009 12:07 am
Thanks a lot, buddies! I made a mistake while calculating in Problem 2, but now i am confident that i did others correctly :)