A coin is flipped 10 times where each flip comes up either heads or tails. How many possible outcomes (a) contain exactly two heads? (b) contain at most three tails? (c) contain the same number of heads and tails?

a. 45

b. 176

c. 252

Step-by-step explanation:

First take into account the concept of combination and permutation:

In the permutation the order is important and it is signed as follows:

P (n, r) = n! / (n - r)!

In the combination the order is NOT important and is signed as follows:

C (n, r) = n! / r! (n - r)!

Now, to start with part a, which corresponds to a combination because the order here is not important. Thus

 n = 10

r = 2

C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

There are 45 possible scenarios.

Part b, would also be a combination, defined as follows

n = 10

r <= 3

Therefore, several cases must be made:

C (10, 0) = 10! / 0! * (10-0)! = 10! / (0! * 10!) = 1

C (10, 1) = 10! / 1! * (10-1)! = 10! / (1! * 9!) = 10

C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

C (10, 3) = 10! / 3! * (10-3)! = 10! / (2! * 7!) = 120

The sum of all these scenarios would give us the number of possible total scenarios:

1 + 10 + 45 + 120 = 176 possible total scenarios.

part c, also corresponds to a combination, and to be equal it must be divided by two since the coin is thrown 10 times, it would be 10/2 = 5, that is our r = 5

Knowing this, the combination formula is applied:

C (10, 5) = 10! / 5! * (10-5)! = 10! / (2! * 5!) = 252

252 possible scenarios to be the same amount of heads and tails.