Suppose in a competition 11 matches are to be played, each having one of 3

distinct outcomes as possibilities. The number of ways one can predict the

outcomes of all 11 matches such that exactly 6 of the predictions turn out to

be correct is

(A) \(^{11}C_{6}\times 2^5\) (B) \(^{11}C_{6}\) (C) \(3^6\)

distinct outcomes as possibilities. The number of ways one can predict the

outcomes of all 11 matches such that exactly 6 of the predictions turn out to

be correct is

(A) \(^{11}C_{6}\times 2^5\) (B) \(^{11}C_{6}\) (C) \(3^6\)

(D) none of the above.

ISI MMA 2015 Question 4

Combinatorics

Easy

Schaum’s Outline of College Algebra

Do you really need a hint? Try it first!

Using the methods of combinatorics we usually count how many ways we can choose some elements out of some total elements, how many ways can we arrange something etc. Here in this problem we have to choose the cases given that condition.

There are different combinations of the different outcomes which can be correct. Basically, we are choosing six games out of 11 and remember that we are not putting any order and we are not required to as well. Now chosen this 6 games we will predict only one option which is the correct one so will multiply with it which will not put any effect on the answer.

You have different choices for the matches which need to be wrong (not 3). Basically, these are the 2 wrong outcomes that have to be predicted. Moreover exactly matches have to be predicted wrong so we will multiply with it. Hence, the final outcome is .

If you need at least outcomes, calculate exactly 6,7,8,9,10 and exactly 11, otherwise you double count some combinations.

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