Noors knows that David, Adam, Jane and Luisa are brothers and sisters. However, she does not know the order in which they were born.
There are 24 different combinations of their birth order.
How many more different combinations would there be if there were five children in the family?
In your first case , there were 4 children and the answer was 4! or 24
So if there are 5 children .... ??
So if there are 5 children there will be 5 or 30 (5*6) is that correct?
4! = 4*3*2*1 = 24
actually the order matters so it is permutations not combinations (google permutations and combinations)
n!
where the ! means "factorial"
n! = n (n-1)(n-2) ....... (1)
5! = 5*4! = 5 * 24 = 120
or if you already know the 24 for four of them, then if you add a fifth, you get five times as many possible orders. That fifth kid could be anywhere in the order, first to fifth
Damon’s answer above is correct for 5 set combination but we have to deduct 24 to get the answer as it says how many more combinations.
So the answer is 120 - 24 = 96.
To figure out the number of different combinations when there are five children in the family, we need to calculate the number of permutations.
In this case, we can use the formula to calculate permutations of n objects taken r at a time, which is given by nPr = n! / (n-r)!
Since there are 5 children in total, we want to calculate the number of permutations taken all at a time. So, n = 5 and r = 5.
Plugging these values into the formula:
5P5 = 5! / (5-5)!
= 5! / 0!
= 5! / 1 (since 0 factorial is defined as 1)
= 5 x 4 x 3 x 2 x 1 / 1
= 120
Therefore, there would be 120 different combinations if there were five children in the family.
Comparing this to the number of combinations when there are four children, we can see that there would be 120 - 24 = 96 more different combinations.