Martin's iPod has 24 songs on it. He wants to make a playlist consisting of three songs to dance to. Let N be the number of different playlists Martin can make. What are the last 3 digits of N?

number of playlists

= C(24,3) = 2024

Why the last 3 digits?
What a strange question.

To find the number of different playlists Martin can make, we need to use combinations.

Since Martin wants to make a playlist of three songs out of the 24 songs on his iPod, we need to calculate the combination or "24 choose 3".

Combination formula: nCk = n! / (k! * (n-k)!)

Using this formula, we can calculate:
24C3 = 24! / (3! * (24-3)!)

Simplifying:

24! = 24 * 23 * 22 * ... * 2 * 1
3! = 3 * 2 * 1
(24-3)! = 21!

24! = 24 * 23 * 22 * 21!

Now we substitute the values:

24C3 = (24 * 23 * 22) / (3 * 2 * 1) * (21!)

Calculating the first part:

24 * 23 * 22 = 12144

Let's simplify the expression further:

24C3 = 12144 / (3 * 2 * 1) * (21!)

Now we calculate (21!):

21! = 21 * 20 * 19 * ... * 2 * 1

Since we only need to find the last 3 digits of N, we can simplify the expression by calculating the factorial modulo 1000.

21! % 1000 = (21 * 20 * 19 * ... * 2 * 1) % 1000

Calculating the second part modulo 1000 might be time-consuming. To simplify the calculation, we can break down 21! into smaller parts and calculate each part modulo 1000.

21! % 1000 = (21 * 20 * 19 * ... * 2 * 1) % 1000
= ((20 % 1000) * (19 % 1000) * ... * (2 % 1000) * (1 % 1000)) % 1000

Calculating each part modulo 1000:

20 % 1000 = 20
19 % 1000 = 19
.
.
.
2 % 1000 = 2
1 % 1000 = 1

Let's substitute these values back into the expression:

21! % 1000 = (20 * 19 * ... * 2 * 1) % 1000

Now, to find the last 3 digits of N, we simplify:

24C3 = (12144 / (3 * 2 * 1) * ((20 * 19 * ... * 2 * 1) % 1000)

Calculating the first part:

12144 / (3 * 2 * 1) = 2024

Substituting back into the expression:

24C3 = 2024 * ((20 * 19 * ... * 2 * 1) % 1000)

Now, we calculate ((20 * 19 * ... * 2 * 1) % 1000) using a calculator or programming language.

After calculating this value, we can then multiply it by 2024 to find the last 3 digits of N.