A team has no order
A coach picks a team of 3 from 8 players. Picking Ann, then Ben, then Cal gives the same team as picking Cal, then Ann, then Ben: nobody on the team has a rank, so only who is in it matters.
A selection in which order does not matter is called a combination. The question "how many teams?" asks for the number of combinations of 3 players from 8.
Count in order, then correct
Counting the picks in order is easy: 8 choices for the first player, 7 for the second and 6 for the third, so 8 × 7 × 6 = 336 ordered picks.
But each team turns up in that count more than once. Ann, Ben and Cal can be picked in 3! = 6 orders: ABC, ACB, BAC, BCA, CAB and CBA. Every team is counted 6 times, so the number of teams is 336 ÷ 6 = 56.
Three picks in order give 336, and each team of 3 was counted once for each of its 3! = 6 orders.
A small case drawn in full
Choose 2 people from 4, called A, B, C and D. In order there are 4 × 3 = 12 picks. As teams, AB and BA are the same, and so are AC and CA, and so on: each team is one of a pair of ordered picks. So there are 12 ÷ 2 = 6 teams: AB, AC, AD, BC, BD and CD.
First pick down the side, second pick across the top. Each team appears twice, once on each side of the empty diagonal, as AB and BA; the 6 shaded squares are one of each.
The formula
The number of ways to choose r things from n, in no order, is written nCr; it is also written as n above r inside one pair of round brackets. Every combination is counted r! times among the nPr ordered picks, so .
With factorials, nCr = n! ÷ (r! × (n − r)!). In practice, write it as a fraction with r factors on top, counting down from n, and r! underneath, then cancel: , and .
Choosing who is left out
Choosing 3 players for the team is the same as choosing the 5 who are not on it. So 8C3 = 8C5 = 56, and in general nCr = nC(n − r). This makes large choices quick: .
There is one way to choose nothing and one way to choose everything, so nC0 = nCn = 1. The formula agrees: nCn = n! ÷ (n! × 0!) = 1, using 0! = 1.
Choosing from two groups
A committee of 2 women and 2 men is chosen from 7 women and 5 men. The women can be chosen in ways and the men in ways.
Both choices are made for every committee, so they multiply: 21 × 10 = 210 committees. Adding them, 21 + 10, would count a choice of women or a choice of men, not both.
The usual mistakes
Leaving the order in. 336 counts ordered picks; a team is the same team in whatever order its members were named, so divide by 3! to get 56.
Dividing by r instead of r!. A team of 3 has 3 × 2 × 1 = 6 orders, not 3, so 336 ÷ 3 = 112 is too many.
Multiplying n by r. A team of 3 from 8 is not 8 × 3 = 24.
Adding the choices from two groups that are both chosen. 7C2 × 5C2 = 210, not 21 + 10 = 31.
A committee, and a box of bulbs
In the first application below, a committee is chosen from two groups, and "at least one man" is counted as every committee minus those with no man.
In the second, three bulbs are taken at once from twelve. Every set of three is equally likely, so a probability is the number of sets that match divided by 12C3 = 220.
Worked example: A Residents' Committee of Four Chosen from Seven Women and Five Men
Question A residents' association chooses a committee of 4 from 7 women and 5 men who have volunteered. (a) How many different committees have exactly 2 women and 2 men? (b) How many different committees include at least one man?
1.(a) Choose 2 of the 7 women: 72 = 7 × 62 × 1 = 21 ways. Choose 2 of the 5 men: 52 = 5 × 42 × 1 = 10 ways.
(a) Choose 2 of the 7 women, 72 = 21 ways, and 2 of the 5 men, 52 = 10 ways. 2.Each choice of women goes with each choice of men, so multiply: 21 × 10 = 210 committees.
Every choice of women goes with every choice of men: 21 × 10 = 210 committees. 3.All committees of 4 from the 12 volunteers: 124 = 12 × 11 × 10 × 94 × 3 × 2 × 1 = 495.
All committees of 4 from the 12 volunteers: 124 = 495. 4.The committees with no man are chosen from the women only: 74 = 7 × 6 × 5 × 44 × 3 × 2 × 1 = 35.
The committees with no man come from the women only: 74 = 35. 5.(b) At least one man is the complement of no man: 495 − 35 = 460 committees. Check by cases: 7351 + 210 + 7153 + 54 = 175 + 210 + 70 + 5 = 460.
(b) At least one man is the complement of no man: 495 − 35 = 460.
Answer: (a) 210 committees; (b) 460 committees
Common mistakes
- Adding 21 + 10 in (a). Choosing the women and choosing the men both happen for every committee, so the counts multiply; adding is for alternatives, one or the other.
- Counting (b) as 5 × 113 = 825: one man first, then any three others. A committee with two men is then counted once for each of its men, so the total is too large.
Worked example: An Inspector Tests Three Bulbs from a Box of Twelve, Four of Which Are Faulty
Question A box holds 12 light bulbs, and 4 of them are faulty. An inspector takes 3 bulbs from the box at random, all at once. (a) Find the probability that none of the three bulbs is faulty. (b) Find the probability that exactly one of the three bulbs is faulty.
1.The number of equally likely selections of 3 bulbs from 12 is 123 = 12 × 11 × 103 × 2 × 1 = 220.
Every set of 3 bulbs is equally likely, and there are 123 = 220 of them. 2.No faulty bulb means all three come from the 8 good bulbs: 83 = 8 × 7 × 63 × 2 × 1 = 56 selections.
No faulty bulb: all three from the 8 good ones, 83 = 56 sets. 3.(a) P(none faulty) = 56220 = 1455.
(a) P(none faulty) = 56220 = 1455. 4.Exactly one faulty: choose the faulty bulb, 41 = 4 ways, and two good bulbs, 82 = 28 ways, giving 4 × 28 = 112 selections.
Exactly one faulty: 41 × 82 = 4 × 28 = 112 sets. 5.(b) P(exactly one faulty) = 112220 = 2855. Check: two or three faulty is 4281 + 43 = 48 + 4 = 52 selections, and 56 + 112 + 52 = 220.
(b) P(exactly one faulty) = 112220 = 2855.
Answer: (a) 1455; (b) 2855
Common mistakes
- Working (a) as (812)3, as if each bulb were put back before the next was taken. The bulbs are taken without replacement, so after one good bulb only 7 of the 11 left are good.
- Forgetting the 41 in (b) and counting 28 selections. Each of the 4 faulty bulbs can be the one in the set, and each gives 28 different sets.