There are 5 members in a group. 3 members are can be chosen to go on a trip. How many different groups of 3 can there be?

Question

Word problem that has me confused

Answer 1

Since the order doesn't matter in this problem, there are 5C3 = 5C2 = 5*4/2 = 10 different groups of 3 that can be picked.

Answer 2

This is the number of combinations of 3 elements out of 5.

Example.If you are given 5 digits 1,2,3,4,5 you can pick up 3 elements in the following ways:

 

123  124  125   234   235   345   134    135   245    145

 

Because the order in which they are picked up doesn't matter (123 is the same collection like 321 or 231)

these are all possible combinations, 10. Robert has shown you the direct way of calculations:

 

5C3 = 5x4x3/1x2x3 = 10.

Answer 3

this is a combination problem

you are choosing 3 people from a group of 5 people

the formula is n!/[(r!)(n-r)!] where n is the number of people(5) and r is how many you are choosing(3)

! means factorial: for example 7!=7*6*5*4*3*2*1

so we have (5*4*3*2*1)/[(3*2*1)(2*1)] because n-r=5-3=2

notice that the numerator and the denominator both have 3*2*1 in them, so they cancel out and you have

(5*4)/(2*1)=20/2=10

if you did not cancel or simplify you would have 120/(6)(2)=120/12=10 again

Answer 4

5*4*3 = number of groups if order matters

3*2*1 = number of ways one group can be ordered

5*4*3 / (3*2*1) = number of groups if order does not matter
= 5*2*2*3*1 / (3*2*1)
= 5*2 = 10

Where do the factorials come from in the formula for mCn?

5*4*3 = 5*4*3*2*1 / (2*1) = 5! / 2!

3*2*1 = 3!

5*4*3 / (3*2*1) = 5! / 2! / 3! = 5! / (2! * 3!) = 5C3 = 5C2