Ten prizes are given to a class with 100 students Each stude

Ten prizes are given to a class with 100 students Each student can receive at most one prize Alice and Bob are two students in the class If the prizes are identical, how many ways are there to distribute the prizes so that either Alice or Bob (or both) receive a prize? If the prizes are different, how many ways are there to distribute the prizes so that either Alice or Bob (or both) receive a prize?

Solution

a. The required case: Either alice, or bob or both should receive a prize.

Number of different scenarios:

1. Only Alice receives a prize, and the other 9 prizes are distributed among 98 students.

2. Only Bob receives a prize, and the other 9 prizes are distributed among 98 students.

3. Both Alice and Bob receives prizes, and remaining 8 prizes are distributed among 98 students.

4. Neither Alice nor Bob receives a prize, and all 10 prizes are distributed among 98 students.

The favorable case can be calculated in 2 different ways:

1. Counting all the favorable cases.

2. Counting all the unfavorable cases, and subtracting them from all possible cases.

(a). The number of possible cases: ¹⁰⁰C₁₀. Here its a combination because, all prizes are identical, and giving prize A to student X, and prize B to student Y is identical to giving prize B to student X, and prize A to student Y.

Number of unfavorable cases: i.e., Neither Alice nor Bob receives a prize: ⁹⁸C₁₀.

And to conunt the favorable cases which is an answer to problem (a) is: subtracting the unfavorable cases from total possibilities. Therefore, the answer is: ¹⁰⁰C₁₀ - ⁹⁸C₁₀ = (100! / (10! * 90!)) - (98! / (10! * 88!)) is the answer.

(b). The number of possible cases: ¹⁰⁰P₁₀. Here its a permutaion because, all prizes are unidentical(different), and giving prize A to student X, and prize B to student Y is not same as giving prize B to student X, and prize A to student Y.

Number of unfavorable cases: i.e., Neither Alice nor Bob receives a prize: ⁹⁸P₁₀.

And to conunt the favorable cases which is an answer to problem (a) is: subtracting the unfavorable cases from total possibilities. Therefore, the answer is: ¹⁰⁰P₁₀ - ⁹⁸P₁₀ = (100! / 90!) - (98! / 88!)) is the answer.