Alexis, Bart, Chuck, and Dariah all called in to a radio show to get free tickets to a concert. In how many different ways could there calls have been received?

Question

I dont understand. It is not making sense.

Answer 1

Hey Emma,

 

This is a permutation question. So basically you want to arrange Alexis, Bart, Chuck and Dariah into an order without repeating them. That means no one person can call twice. 

 

So the hard way of doing this is doing each possibility by hand for example three possibilities are:

 

alexis, bart, chuck, dariah
alexis, bart, dariah, chuck
alexis, chuck, bart, dariah

But the mathematical way of doing this is first recognizing that there is no repetition so no person calls twice. Next there are four places people can call in, 1st, 2nd, 3rd and 4th. And there are 4 people. 

 

So the number of people that can call in first is 4, the number of people that can in second is 3 because the person that called in first can't call again. So the number of people that can call in third is 2 and then only one person can call in last.

 

So the answer is 4x3x2x1 = 24 different ways. This is also written as 4! which just means multiply a series of descending natural numbers. Examples:

4! = 4 × 3 × 2 × 1 = 24
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

1! = 1

Answer 2

Let's do he case when Alexis calls first.  After Alexis calls, the order of the remaining calls are:

1. Bart Chuck Dariah
2. Bart Dariah Chuck
3. Chuck Bart Dariah
4. Chuck Dariah Bart
5. Dariah Bart Chuck
6. Dariah Chuck Bart

 

So there are 6 different ways the calls can be received when Alexis calls first.  Similarly, there are 6 ways when Bart calls first, 6 more when Chuck calls first, and 6 more when Dariah calls first.  The total number of ways is 6 + 6 +6 + 6 = 24 ways the calls can be received.  This is actually a permutation because the order that the calls are received matters, but we cannot repeat an order.  The formula for the non-repeating permutations is:

 

      n!/(n-r)!  where n is the number of callers and r is the calls; so n=4 and r = 4

 

      n!/(n-r)! = 4!/(4-4)! = 4!/0! = 4*3*2*1/1 = 24  (Note that 0! = 1)

Answer 3

Let's ignore the fact that anyone else could have called in besides these four and that then order determines who wins the free tickets. Then we're talking about the total number of unique ways these four people can call (with a specific order) to calculate the odds that each person wins. Because order matters, this means we're dealing with a permutation, not a combination. There are four people involved so 4!=24.

 

Hope that helps!