A. Counting and Probability · Permutations and Combinations

Combination Problems: Groups, "At Least" and Cards

A2.2 A2.3About 70 minutes

Learning goals

I can choose from two or more groups by multiplying combinations.
I can solve "at least" and "at most" problems with cases or with the complement.
I can count selections that must include or exclude particular objects.
I can count hands of cards and subsets of any size.

Words to know

at least k
k or more. "At least 2 teachers" means 2, 3, 4, … teachers.

at most k
k or fewer. "At most 1 woman" means 0 or 1 woman.

standard deck
52 playing cards in 4 suits (hearts and diamonds are red; clubs and spades are black). Each suit has 13 ranks: ace, 2 to 10, jack, queen, king.

hand
A set of cards dealt to a player. The order in which the cards arrive does not matter, so hands are combinations.

subset
Any selection from a set, of any size from none to all. A set of n objects has 2ⁿ subsets.

Real selections have rules

A school is forming a 5-person committee to plan the winter carnival, choosing from 6 teachers and 8 students. The principal might want "exactly 2 teachers", "at least 3 teachers", or "Ms. Roy must be on it". Each rule changes the count. This lesson combines the combination formula with the additive and multiplicative principles from lessons 1 and 2.

With no rules at all, there are C(14, 5) = 2002 possible committees.

Choosing from separate groups

When a selection takes a certain number from each of several groups, choose from each group separately and multiply (the multiplicative principle: teachers and students).

Choose the teachers and the students separately, then multiply. The shaded people are one of the 840 possible committees.

Worked example

Example 1: Exactly 2 teachers

How many 5-person committees from 6 teachers and 8 students have exactly 2 teachers?

Exactly 2 teachers means exactly 3 students.
Teachers: C(6, 2) = 15. Students: C(8, 3) = 56.
Multiply: 15 × 56 = 840.

840 committees.

Worked example

Example 2: At least 3 teachers (cases)

How many 5-person committees from 6 teachers and 8 students have at least 3 teachers?

"At least 3" means 3, 4 or 5 teachers. These cases cannot overlap, so count each and add.
3 teachers, 2 students: C(6, 3) × C(8, 2) = 20 × 28 = 560.
4 teachers, 1 student: C(6, 4) × C(8, 1) = 15 × 8 = 120.
5 teachers, 0 students: C(6, 5) × C(8, 0) = 6 × 1 = 6.
Add: 560 + 120 + 6 = 686.

686 committees.

"At least 3 teachers" splits into three non-overlapping cases. Multiply within each case, then add the cases.

Worked example

Example 3: At least one teacher (complement)

How many 5-person committees from 6 teachers and 8 students include at least one teacher?

Directly, this would need 5 cases (1, 2, 3, 4 or 5 teachers).
The complement is "no teachers": all 5 are students: C(8, 5) = 56.
Subtract from the total: 2002 − 56 = 1946.

1946 committees.

A tempting mistake

For "at least one teacher", some students try: choose 1 teacher (6 ways), then any 4 of the remaining 13 people (C(13, 4) = 715), giving 6 × 715 = 4290. That is more than the total of 2002, so it must be wrong. A committee with two teachers, T1 and T2, is counted twice: once with T1 as "the chosen teacher" and once with T2. Use cases or the complement instead.

Worked example

Example 4: Including or excluding a particular person

How many 5-person committees from the 14 people (a) must include Ms. Roy, (b) must exclude Ms. Roy?

(a) Put Ms. Roy on the committee. Choose the other 4 from the remaining 13: C(13, 4) = 715.
(b) Remove Ms. Roy. Choose all 5 from the other 13: C(13, 5) = 1287.
Check: every committee either includes her or does not, so 715 + 1287 should be the total: 2002 ✓.

(a) 715; (b) 1287.

Hands of cards

A 5-card hand from a standard deck is a combination: C(52, 5) = 2 598 960 hands. To count hands of a particular type, choose from each suit or rank separately.

Exactly 3 hearts: 3 of the 13 hearts and 2 of the 39 non-hearts: C(13, 3) × C(39, 2) = 286 × 741 = 211 926.
All the same suit: choose the suit (4 ways), then 5 of its 13 cards: 4 × C(13, 5) = 4 × 1287 = 5148.

Any number of objects

If you may choose any number of n objects (including none), each object is either in or out: 2 choices each, so 2ⁿ selections. A pizza with any of 8 toppings: 2⁸ = 256 possibilities, including the plain pizza. If at least one topping is required: 2⁸ − 1 = 255.

Just for fun

Of the 2 598 960 possible five-card hands, only 3744 are full houses, about 1 in 694. So if a friend says they "always" get one, the math respectfully disagrees.

Hands-on activity

A seven-card mini-deck

You need: 4 hearts and 3 spades from a deck of cards (7 cards in total).

Predict how many different 3-card hands can be made from these 7 cards.
Predict how many of those hands have exactly 2 hearts. Then list them by pairing each of the C(4, 2) heart pairs with each spade.
Predict, using the complement, how many hands have at least one spade.
Deal 3 cards at random 20 times, shuffling between deals. Tally how often you get exactly 2 hearts. Compare your fraction with your count divided by the total.

What to look for: C(7, 3) = 35 hands. Exactly 2 hearts: C(4, 2) × C(3, 1) = 6 × 3 = 18. At least one spade: 35 − C(4, 3) = 35 − 4 = 31. Your experimental fraction should be roughly 18/35 ≈ 0.51; with only 20 deals it will vary. Turning counts into probabilities is the subject of lessons 13 and 14.

Your turn

A 4-person committee is chosen from 5 boys and 6 girls. How many committees have exactly 2 girls?

How many of those 4-person committees have at least one girl?

How many 5-card hands contain exactly 2 aces?

How many different 5-card hands can be dealt from a standard deck?

How many 5-card hands consist only of face cards (jacks, queens and kings)?

A team of 6 is chosen from 10 players and must include the captain. How many teams?

An ice cream sundae can have any number of 7 toppings (including none). How many sundaes are possible?

More practice: warm-up

Choose 2 of 5 apples and 3 of 4 oranges. How many selections?

A 3-person committee is chosen from 7 people and must include Sam. How many committees?

A 3-person committee is chosen from 7 people and must not include Sam. How many committees?

How many subsets does the set {a, b, c, d} have?

How many 2-card hands can be dealt from a standard deck?

More practice: on your own

A 5-person committee is chosen from 7 men and 6 women. How many committees have at least 3 women?

For the same group, how many committees have at most 1 woman?

How many 5-card hands contain at least one ace?

A full house is 3 cards of one rank and 2 of another (for example three 7s and two kings). How many full houses are there?

Jordan counts hands with at least one ace as C(4, 1) × C(51, 4) = 999 600. Explain the error.

Challenge

In how many ways can 12 students be split into 3 project groups of 4, if the groups are not labelled?

How many 5-card hands contain exactly two pairs (for example two 9s, two jacks and a 4)?

A pizza can have any number of 10 toppings, but must have at least 2. How many pizzas are possible?

Check your understanding

Quiz

From 6 teachers and 8 students, how many 5-person committees have exactly 2 teachers?

From 6 teachers and 8 students, how many 5-person committees include at least one teacher?

What is usually the fastest way to count "at least one"?

How many 5-card hands can be dealt from a standard 52-card deck?

A pizza can have any number of 8 toppings, including none. How many pizzas are possible?

A 4-person team is chosen from 10 people and must include Kai. How many teams?

For parents and teachers

Big idea (A2.2, A2.3): multiply combinations when choosing from separate groups, add cases for "at least" or "at most", and use the complement when there are many cases. Card hands are combinations because the order of dealing does not matter.

How to help: Before calculating, ask your teen to say what the cases are ("at least 3 means 3, 4 or 5"). A sanity check is powerful: no answer can be bigger than the total with no conditions.

Common difficulties:

The "choose one first, then the rest" error for "at least one", which overcounts.
Forgetting a case (for example the 5-teacher case).
Mixing up the number in a group with the number chosen.
Not knowing the deck: 52 cards, 4 suits, 13 ranks, 12 face cards. A deck at home is the best reference.