# can someone help?

Suppose there are 10 friends going to the fair. One attraction allows only three of the students to enter at a time. How many different groups of three students can be selected for the attraction?

may someone help?

### 3 Answers

- PuzzlingLv 76 months agoFavorite Answer
There are 10 friends and you are choosing 3 of them.

Let's start by picking them in order:

1st person = 10 choices

2nd person = 9 choices

3rd person = 8 choices

10 * 9 * 8 = 720 permutations of 10 people taken 3 at a time.

However, we don't care about the order the people are chosen. If I pick Alan, Becky and Chuck, that's not any different than picking Chuck, Alan and Becky, or any of the other orders of these 3 people. With 3 people, there are 3! = 3 * 2 * 1 = 6 ways to order them.

To account for the different orders and counting them as only 1 combination, we need to divide our permutation answer by 6.

720 / 6 = 120 ways

There is a formula for the number of combinations of n people chosen k at a time.

C(n,k) = n! / (k! (n-k)!)

C(10,3) = 10! / (3! 7!) = 120

I have trouble remembering the formula, so I just think of C(10,3) as a fraction first:

10

---

3

Then I multiply top and bottom by the next smaller number. I do that until I get to 1 on the bottom.

10 * 9 * 8

-------------

3 * 2 * 1

After that, I look for ways to cancel all the numbers on the bottom. For example, I can cancel a 3 from 3 and 9. And I can cancel a 2 from 2 and 8.

10 * 3 * 4

-------------

1 * 1 * 1

Once the denominator is all ones, I just multiply what's left in the numerator.

10 * 3 * 4 = 120 ways

If you want to check your work, just type "10 choose 3" into Google to verify you have the right answer.

Answer:

120 combination

Source(s): https://www.google.com/search?q=10+choose+3- Login to reply the answers

- Anonymous6 months ago
Thus must be set up in combinatorial form of (10 * 9 * 8) to create set pairing.

Simple application of counting technique.

- Login to reply the answers

- GregLv 76 months ago
ABCDEFGHIJ = 10

ABC, ABD, ABE, ABF, ABG, ABH, ABI, ABJ, BCD, BCE, BCF, BCG, BCH, BCI, BCJ,

CDE, CDF, CDG, CDH, CDI, CDJ, DEF, DEG, DEH, DEI, DEJ, EFG, EFH, EFI, EFJ,

FGH, FGI, FGJ, GHI, GHJ, HIJ count them or use 8+7+6+5+4+3+2+1 = 36

I bet you were "taught" an equation to use.

- nbsaleLv 66 months agoReport
Not right. For example: Where is ACE? Do you see what you did wrong? There are too many to take an approach of listing all possible triplets.

- Login to reply the answers