Answer:
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.
