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Combination!

by gmat_perfect » Sat Nov 06, 2010 12:57 am
There are 6 food items P, Q, R, S, T, U. If we want to make mixture of different types of food items, how many combination is possible if the items T and U are not allowed to mix together?

Please solve with explanation.

Thanks.
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by Rahul@gurome » Sat Nov 06, 2010 6:45 am
2 food items can be chosen in 6C2 = 6!/(2!)(4!) = 15 ways, T and U can occur together in 1 way, so 2 food items such that T and U doesn't occur together can be chosen in 15 - 1 = 14 ways
3 food items can be chosen in 6C3 = 6!/(3!)(3!) = 20 ways, T and U can occur together in 4 ways, so 3 food items such that T and U doesn't occur together can be chosen in 20 - 4 = 16 ways
4 food items can be chosen in 6C4 = 6!/(4!)(2!) = 15 ways, T and U can occur together in 6 ways, so 4 food items such that T and U doesn't occur together can be chosen in 15 - 6 = 9 ways
5 food items can be chosen in 6C5 = 2 ways

Therefore, required number of combinations = 14 + 16 + 9 + 2 = 41 ways
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by gmat_perfect » Sat Nov 06, 2010 10:37 am
Rahul@gurome wrote:2 food items can be chosen in 6C2 = 6!/(2!)(4!) = 15 ways, T and U can occur together in 1 way, so 2 food items such that T and U doesn't occur together can be chosen in 15 - 1 = 14 ways
3 food items can be chosen in 6C3 = 6!/(3!)(3!) = 20 ways, T and U can occur together in 4 ways, so 3 food items such that T and U doesn't occur together can be chosen in 20 - 4 = 16 ways
4 food items can be chosen in 6C4 = 6!/(4!)(2!) = 15 ways, T and U can occur together in 6 ways, so 4 food items such that T and U doesn't occur together can be chosen in 15 - 6 = 9 ways
5 food items can be chosen in 6C5 = 2 ways

Therefore, required number of combinations = 14 + 16 + 9 + 2 = 41 ways
Thanks for such a nice explanation.

I could not get one point.

In case of 3 items, how have you got 4?
In case of 4 items, how have you got 6?

Thanks.
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by goyalsau » Sat Nov 06, 2010 8:59 pm
I am not Rahul, But I think its worthing trying explain, Because i m very Poor in SC and do not like to visit that section of the test at all, but at times when i try to study your post are very helpful so, I think I must try to help you understand.

We have 6 Items in all,

For two items together, 6C2 ways,
we have to subtract the case where T and U are together, so 2C2 ways, , ( That is because we are only considering T and U { WE have to determine How many ways then can be written together } that is 1 way ,

Now Three items together, 6C3 ways
we have to subtract the case where T and U are together, with any other element ( Now its like this T U one more from the 4 elements that are left { we can choose one element from 4 in 4C1 ways } that will be 4 )

Now Four Items together , 6C4 ways,
we have to subtract the case where T and U are together, with two other element ( Now its like T U and two more elements { we can choose them in 4C2 ways, } that will be 6 )

I think by now you must have understood the concept ,

I would like to thank you for those wonder full SC post of yours and Please if you can suggest me something, I am Very VERY VERy........................... infinite Very............ Poor in SC,
I just hate it,
I don't want to study it at all, but in Verbal it has more than 13 questions , So what should i do, please PM me buddy..........
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by Rahul@gurome » Sun Nov 07, 2010 4:05 am
gmat_perfect wrote:
Thanks for such a nice explanation.

I could not get one point.

In case of 3 items, how have you got 4?
In case of 4 items, how have you got 6?

Thanks.
PTU, QTU, RTU, STU are 4 ways where T and U occur together.
PQTU, PRTU, PSTU, QRTU, QSTU, RSTU are 6 ways where T and U occur together.

Hope that helps.
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