How many ways can three prizes be given away to 12 students when each student is eligible for all the prizes?

In how many ways can 3 prizes be given away to 12 students when each student is eligible for all the prizes? 1234.

How many ways can we select five door prizes from ten different ones and distribute them among five people?

How many ways can five different door prizes be distributed among five people? There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 · 4 · 3 · 2 · 1 = 120 ways.

How many words of 4 letters with or without meaning be made from the letters of the word number when repetition of letters is not allowed?

Therefore, the number of four-letter words that can be formed is 5040.

How many ways can 3 prizes be distributed?

∴ The number of ways in which all the 3 prizes can be given = 4 × 3 × 2 = 24.

How many ways can 5 prizes?

The answer says 1024, which I guess is coming from 45. So its like “4 students fit into each prize” + repeats allowed + 5 prizes =45=1024.

How many ways can 1st 2nd and 3rd place prizes be awarded for a 10 contestant pageant you can’t win twice?

For each of 10, there are 9. For each of those, there are 8. So we end up with Robert’s simple calculation of 10 x 9 x 8 = 720 ways.

How many different combinations of three can we have on the podium?

3!)= 56 ways to select the three that will be on the podium.

How many 3 letter words can be formed using the letters of the word time?

number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.

How many prizes will be given to the students at once?

Not all students are going to win the prize, and only at most 3 out of 12 will be awarded. Therefore, only 3 prizes will be given to the students at a time, and as in the formula, n^r, n will be 12 and r will be 3.

How many ways can five different door prizes be distributed?

How many ways can five different door prizes be distributed among five people? There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 · 4 · 3 · 2 · 1 = 120 ways. Now we will consider some slightly different examples. Example 27

How many different medals can be awarded to the 8 runners?

Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 · 7 · 6 = 336 ways the three medals can be awarded to the 8 runners.