7.6 Combinations Involving Several Sets
Learning Objectives
By the end of this section, you will be able to:
- Count the number of items selected when the selection is made from more than one set.
- Count the number of items selected when there are restrictions on how many come from each set.
In Section 7.5 we solved the basic combination problem of choosing \(r\) objects from a single pool of \(n\). Real problems are rarely that clean — committees pull from separate pools of men and women, hands of cards pull from four suits, playlists pull from different genres. This section shows how to combine \({}_n\mathrm{C}_r\) with the multiplication axiom to count selections drawn from several sets.
7.6.1 Selecting from Several Sets
When a selection draws a fixed number of items from each of several distinct pools, we compute one combination per pool and multiply. This is just the multiplication axiom from Section 7.2 applied to each \({}_n\mathrm{C}_r\) factor.
A committee with 2 men picked from 4 and 3 women picked from 4 is really two independent selections glued together. The men's side offers \({}_4\mathrm{C}_2 = 6\) options and the women's side offers \({}_4\mathrm{C}_3 = 4\) options. Every men's pick can be paired with every women's pick, so the multiplication axiom gives \(6 \cdot 4 = 24\) total committees. Whenever a problem says "pick \(r_1\) from this set AND \(r_2\) from that set," the word and becomes multiplication.
Re-read any committee or playlist problem and circle every phrase of the form "\(r\) from the \(n\) [things]." Each circle becomes one \({}_n\mathrm{C}_r\) factor. Multiply the circles together and you are done.
Source: Applied Finite Mathematics
When a selection draws \(r_i\) objects from the \(i\)-th of several disjoint pools of sizes \(n_1, n_2, \ldots, n_k\), the total number of such selections is the product
$${}_{n_1}\mathrm{C}_{r_1} \cdot {}_{n_2}\mathrm{C}_{r_2} \cdots {}_{n_k}\mathrm{C}_{r_k}.$$Source: Applied Finite Mathematics
How many five-person committees consisting of 2 men and 3 women can be chosen from a total of 4 men and 4 women?
Solution
Step 1 — Identify the two pools. We pull \(2\) men from a pool of \(4\) men, and \(3\) women from a pool of \(4\) women. The two selections are independent.
Step 2 — Count each pool with a combination.
$$\text{Men: } {}_4\mathrm{C}_2 = \dfrac{4!}{2!\,2!} = 6.$$ $$\text{Women: } {}_4\mathrm{C}_3 = \dfrac{4!}{3!\,1!} = 4.$$Step 3 — Apply the multiplication axiom. Every 2-man committee can be paired with every 3-woman committee.
$$6 \cdot 4 = 24.$$Answer: \(24\) five-person committees.
Source: Applied Finite Mathematics
How many five-letter word sequences consisting of 2 vowels and 3 consonants can be formed from the letters of the word INTRODUCE?
Solution
Step 1 — Count vowels and consonants in INTRODUCE. The nine letters are I, N, T, R, O, D, U, C, E: that is \(4\) vowels (I, O, U, E) and \(5\) consonants (N, T, R, D, C).
Step 2 — Choose the letters. Pick \(2\) of the \(4\) vowels and \(3\) of the \(5\) consonants.
$${}_4\mathrm{C}_2 \cdot {}_5\mathrm{C}_3 = 6 \cdot 10 = 60.$$Step 3 — Arrange the chosen letters. Each 5-letter group can be ordered in \(5!\) different sequences.
$$60 \cdot 5! = 60 \cdot 120 = 7{,}200.$$Answer: \(7{,}200\) word sequences.
Source: Applied Finite Mathematics
A club has 5 men and 7 women. In how many ways can a committee of 2 men and 2 women be chosen?
Solution
Step 1 — One combination per pool.
$${}_5\mathrm{C}_2 \cdot {}_7\mathrm{C}_2 = 10 \cdot 21 = 210.$$Answer: \(210\) committees.
7.6.2 Combinations with Restrictions
Some problems add words like all, none, or at least. These translate into a short list of cases that we handle separately, then add together. The tools are still combinations and the multiplication axiom — we just use addition to combine disjoint cases.
Combinations cannot directly count "at least three" in one formula. Instead, split the requirement into exact cases — exactly three OR exactly four — and add. As long as the cases cannot overlap, adding is safe. This is the same addition principle you met when counting events that can happen in alternative ways.
"No freshmen on the committee" is the same as "every committee member is a non-freshman." Instead of subtracting, just shrink the pool: pick all members from the non-freshman students and you automatically exclude the freshmen.
Source: Applied Finite Mathematics
A high school club consists of 4 freshmen, 5 sophomores, 5 juniors, and 6 seniors. How many ways can a committee of 4 people be chosen that includes
a) One student from each class?
b) All juniors?
c) Two freshmen and 2 seniors?
d) No freshmen?
e) At least three seniors?
Solution
Part (a) — One from each class. Pick \(1\) from each pool and multiply.
$${}_4\mathrm{C}_1 \cdot {}_5\mathrm{C}_1 \cdot {}_5\mathrm{C}_1 \cdot {}_6\mathrm{C}_1 = 4 \cdot 5 \cdot 5 \cdot 6 = 600.$$Part (b) — All juniors. Pick all \(4\) members from the \(5\) juniors and nothing from the rest.
$${}_5\mathrm{C}_4 = 5.$$Part (c) — Two freshmen and two seniors. Pick \(2\) of the \(4\) freshmen and \(2\) of the \(6\) seniors.
$${}_4\mathrm{C}_2 \cdot {}_6\mathrm{C}_2 = 6 \cdot 15 = 90.$$Part (d) — No freshmen. The committee must come entirely from the remaining \(5 + 5 + 6 = 16\) non-freshman students.
$${}_{16}\mathrm{C}_4 = \dfrac{16!}{4!\,12!} = 1{,}820.$$Part (e) — At least three seniors. Split into "exactly 3 seniors" and "exactly 4 seniors." With \(6\) seniors and \(4 + 5 + 5 = 14\) non-seniors:
$$\underbrace{{}_6\mathrm{C}_3 \cdot {}_{14}\mathrm{C}_1}_{\text{3 seniors}} + \underbrace{{}_6\mathrm{C}_4}_{\text{4 seniors}} = 20 \cdot 14 + 15 = 280 + 15 = 295.$$Answer: (a) \(600\), (b) \(5\), (c) \(90\), (d) \(1{,}820\), (e) \(295\).
Source: Applied Finite Mathematics
From a group of 4 men and 6 women, how many 4-person committees contain at least 3 women?
Solution
Step 1 — Split into cases. "At least 3 women" means exactly 3 women or exactly 4 women.
Step 2 — Count each case.
$${}_6\mathrm{C}_3 \cdot {}_4\mathrm{C}_1 + {}_6\mathrm{C}_4 = 20 \cdot 4 + 15 = 80 + 15 = 95.$$Answer: \(95\) committees.
7.6.3 Combinations Applied to a Deck of Cards
Card-hand problems are a classic testing ground for combinations because the deck has a rich structure — \(52\) cards split into \(4\) suits of \(13\) denominations — and a single hand can require picking from several pieces of that structure at once. The method is exactly the same as before: one \({}_n\mathrm{C}_r\) factor per "from" phrase, multiplied together, with cases added when the problem says at least or or.
Source: Applied Finite Mathematics
A standard deck of playing cards has \(52\) cards consisting of \(4\) suits each with \(13\) cards. In how many different ways can a 5-card hand consisting of four cards of one suit and one of another be drawn?
Solution
Step 1 — Choose the first suit. Any of the \(4\) suits can supply the four-of-a-suit.
$${}_4\mathrm{C}_1 = 4.$$Step 2 — Choose 4 cards from that suit. Each suit has \(13\) cards.
$${}_{13}\mathrm{C}_4 = 715.$$Step 3 — Choose the other suit. One of the remaining \(3\) suits supplies the single card.
$${}_3\mathrm{C}_1 = 3.$$Step 4 — Choose 1 card from that suit.
$${}_{13}\mathrm{C}_1 = 13.$$Step 5 — Multiply.
$$4 \cdot 715 \cdot 3 \cdot 13 = 111{,}540.$$Answer: \(111{,}540\) five-card hands.
A Standard Deck of 52 Playing Cards
A standard deck of \(52\) playing cards has \(4\) suits with \(13\) cards in each suit. The suits are diamonds (\(\diamondsuit\), red), hearts (\(\heartsuit\), red), spades (\(\spadesuit\), black), and clubs (\(\clubsuit\), black). Each suit contains \(13\) denominations: the numbers \(2, 3, 4, \ldots, 10\) and the four special cards Jack (J), Queen (Q), King (K), and Ace (A). The Jack, Queen, and King are called face cards because they have pictures on them. A standard deck therefore has \(3 \cdot 4 = 12\) face cards and \(4 \cdot 4 = 16\) cards of any given non-numeric denomination across the suits. Many examples and homework problems in this book refer to a standard deck of \(52\) playing cards, so keep this structure handy.
Source: Applied Finite Mathematics
How many 5-card hands from a standard deck contain exactly 3 hearts and 2 spades?
Solution
Step 1 — Pick the hearts and the spades separately.
$${}_{13}\mathrm{C}_3 \cdot {}_{13}\mathrm{C}_2 = 286 \cdot 78 = 22{,}308.$$Answer: \(22{,}308\) hands.
Summary
Selecting from Several Sets
When a selection pulls \(r_i\) items from the \(i\)-th of several distinct pools of size \(n_i\), the number of ways to make the selection is
$${}_{n_1}\mathrm{C}_{r_1} \cdot {}_{n_2}\mathrm{C}_{r_2} \cdots {}_{n_k}\mathrm{C}_{r_k}.$$
Restrictions: "At least" / "At most"
Split the requirement into disjoint exact-count cases, count each case with the selecting-from-several-sets rule above, then add the counts.
"None of this type"
A restriction "none from pool \(X\)" is the same as "every member comes from the pools other than \(X\)." Shrink the available pool and apply a single combination.
Problem Set 7.6
Source: Applied Finite Mathematics
Problems 7.6.1–7.6.6: Do the following problems using combinations drawn from several sets.
Problem 1. How many 5-person committees consisting of three boys and two girls can be chosen from a group of four boys and four girls?
Problem 1 Solution
Step 1 — Identify the pools. Pick \(3\) boys from \(4\) and \(2\) girls from \(4\).
Step 2 — Multiply the combinations.
$${}_4\mathrm{C}_3 \cdot {}_4\mathrm{C}_2 = 4 \cdot 6 = 24.$$Answer: \(24\) committees.
Problem 2. A club has 4 men, 5 women, 8 boys and 10 girls as members. In how many ways can a group of 2 men, 3 women, 4 boys and 4 girls be chosen?
Problem 2 Solution
Step 1 — One combination per pool. Pick \(2\) men from \(4\), \(3\) women from \(5\), \(4\) boys from \(8\), \(4\) girls from \(10\).
Step 2 — Evaluate each combination.
$${}_4\mathrm{C}_2 = 6, \quad {}_5\mathrm{C}_3 = 10, \quad {}_8\mathrm{C}_4 = 70, \quad {}_{10}\mathrm{C}_4 = 210.$$Step 3 — Multiply.
$$6 \cdot 10 \cdot 70 \cdot 210 = 882{,}000.$$Answer: \(882{,}000\) groups.
Problem 3. How many 4-person committees chosen from 4 men and 6 women will have at least 3 men?
Problem 3 Solution
Step 1 — Split "at least 3 men" into exact cases. With \(4\) men and \(6\) women, "at least \(3\) men" means exactly \(3\) men or all \(4\) men.
Step 2 — Count each case.
$${}_4\mathrm{C}_3 \cdot {}_6\mathrm{C}_1 + {}_4\mathrm{C}_4 = 4 \cdot 6 + 1 = 25.$$Answer: \(25\) committees.
Problem 4. A batch contains 10 transistors of which three are defective. If three are chosen, in how many ways can they be selected with two defective?
Problem 4 Solution
Step 1 — Identify the pools. Of the \(10\) transistors, \(3\) are defective and \(7\) are non-defective. We want exactly \(2\) defective and \(1\) non-defective.
Step 2 — Multiply the combinations.
$${}_3\mathrm{C}_2 \cdot {}_7\mathrm{C}_1 = 3 \cdot 7 = 21.$$Answer: \(21\) ways.
Problem 5. In how many ways can five counters labeled A, B, C, D and E at a store be staffed by two men and three women chosen from a group of four men and six women?
Problem 5 Solution
Step 1 — Choose the people. Pick \(2\) men from \(4\) and \(3\) women from \(6\).
$${}_4\mathrm{C}_2 \cdot {}_6\mathrm{C}_3 = 6 \cdot 20 = 120.$$Step 2 — Assign them to the five labeled counters. Counters A, B, C, D, E are distinguishable, so the \(5\) selected people can be arranged in \(5! = 120\) ways.
Step 3 — Multiply.
$$120 \cdot 120 = 14{,}400.$$Answer: \(14{,}400\) staffings.
Problem 6. How many 4-letter word sequences consisting of two vowels and two consonants can be made from the letters of the word PHOENIX if no letter is repeated?
Problems 7.6.7–7.6.12: Three marbles are chosen from an urn that contains 5 red, 4 white, and 3 blue marbles. How many samples of the following type are possible?
Problem 6 Solution
Step 1 — Count vowels and consonants in PHOENIX. The letters are P, H, O, E, N, I, X: \(3\) vowels (O, E, I) and \(4\) consonants (P, H, N, X).
Step 2 — Choose the letters. Pick \(2\) of the \(3\) vowels and \(2\) of the \(4\) consonants.
$${}_3\mathrm{C}_2 \cdot {}_4\mathrm{C}_2 = 3 \cdot 6 = 18.$$Step 3 — Arrange each 4-letter group. \(4! = 24\).
Step 4 — Multiply.
$$18 \cdot 24 = 432.$$Answer: \(432\) word sequences.
Problem 7. All three white.
Problem 7 Solution
Step 1 — Choose all 3 from the 4 white marbles.
$${}_4\mathrm{C}_3 = 4.$$Answer: \(4\) samples.
Problem 8. Two blue and one white.
Problem 8 Solution
Step 1 — Two blue from 3, one white from 4.
$${}_3\mathrm{C}_2 \cdot {}_4\mathrm{C}_1 = 3 \cdot 4 = 12.$$Answer: \(12\) samples.
Problem 9. One of each color.
Problem 9 Solution
Step 1 — One of each color. \(5\) red, \(4\) white, \(3\) blue.
$${}_5\mathrm{C}_1 \cdot {}_4\mathrm{C}_1 \cdot {}_3\mathrm{C}_1 = 5 \cdot 4 \cdot 3 = 60.$$Answer: \(60\) samples.
Problem 10. All three of the same color.
Problem 10 Solution
Step 1 — Three cases, one per color. All three red, all three white, or all three blue — these are disjoint, so add.
$${}_5\mathrm{C}_3 + {}_4\mathrm{C}_3 + {}_3\mathrm{C}_3 = 10 + 4 + 1 = 15.$$Answer: \(15\) samples.
Problem 11. At least two red.
Problem 11 Solution
Step 1 — Split "at least 2 red" into exact cases. Exactly \(2\) red (and \(1\) non-red) or exactly \(3\) red. Non-red marbles: \(4 + 3 = 7\).
Step 2 — Count each case.
$${}_5\mathrm{C}_2 \cdot {}_7\mathrm{C}_1 + {}_5\mathrm{C}_3 = 10 \cdot 7 + 10 = 70 + 10 = 80.$$Answer: \(80\) samples.
Problem 12. None red.
Problems 7.6.13–7.6.18: Five coins are chosen from a bag that contains 4 dimes, 5 nickels, and 6 pennies. How many samples of five coins of the following types are possible?
Problem 12 Solution
Step 1 — "None red" = choose all 3 from the non-red marbles. There are \(4 + 3 = 7\) non-red marbles.
$${}_7\mathrm{C}_3 = 35.$$Answer: \(35\) samples.
Problem 13. At least four nickels.
Problem 13 Solution
Step 1 — Split "at least 4 nickels" into exact cases. There are \(5\) nickels and \(4 + 6 = 10\) non-nickels. Exactly \(4\) nickels (and \(1\) non-nickel) or exactly \(5\) nickels.
Step 2 — Count each case.
$${}_5\mathrm{C}_4 \cdot {}_{10}\mathrm{C}_1 + {}_5\mathrm{C}_5 = 5 \cdot 10 + 1 = 51.$$Answer: \(51\) samples.
Problem 14. No pennies.
Problem 14 Solution
Step 1 — "No pennies" = choose all 5 from dimes and nickels. There are \(4 + 5 = 9\) non-penny coins.
$${}_9\mathrm{C}_5 = 126.$$Answer: \(126\) samples.
Problem 15. Five of a kind.
Problem 15 Solution
Step 1 — Check which denominations have enough coins for "all 5". Dimes: only \(4\) available, impossible. Nickels: \(5\) available, \({}_5\mathrm{C}_5 = 1\). Pennies: \(6\) available, \({}_6\mathrm{C}_5 = 6\).
Step 2 — Add the two feasible cases.
$$0 + 1 + 6 = 7.$$Answer: \(7\) samples.
Problem 16. Four of a kind.
Problem 16 Solution
Step 1 — Three cases: four dimes, four nickels, or four pennies, plus one coin of any other type.
- Four dimes: \({}_4\mathrm{C}_4 \cdot {}_{11}\mathrm{C}_1 = 1 \cdot 11 = 11\). (The \(11\) non-dimes are \(5\) nickels + \(6\) pennies.) - Four nickels: \({}_5\mathrm{C}_4 \cdot {}_{10}\mathrm{C}_1 = 5 \cdot 10 = 50\). (The \(10\) non-nickels are \(4\) dimes + \(6\) pennies.) - Four pennies: \({}_6\mathrm{C}_4 \cdot {}_9\mathrm{C}_1 = 15 \cdot 9 = 135\). (The \(9\) non-pennies are \(4\) dimes + \(5\) nickels.)
Step 2 — Add the disjoint cases.
$$11 + 50 + 135 = 196.$$Answer: \(196\) samples.
Problem 17. Two of one kind and two of another kind.
Problem 17 Solution
Step 1 — Identify the structure. "Two of one kind and two of another kind" accounts for \(4\) coins; the fifth coin must be of the remaining third kind (otherwise we would have "two, two, and one more of a kind already counted," which contradicts the wording).
Step 2 — Enumerate the three kind-pair choices.
- Doubles dimes + nickels, third coin a penny: \({}_4\mathrm{C}_2 \cdot {}_5\mathrm{C}_2 \cdot {}_6\mathrm{C}_1 = 6 \cdot 10 \cdot 6 = 360\). - Doubles dimes + pennies, third coin a nickel: \({}_4\mathrm{C}_2 \cdot {}_6\mathrm{C}_2 \cdot {}_5\mathrm{C}_1 = 6 \cdot 15 \cdot 5 = 450\). - Doubles nickels + pennies, third coin a dime: \({}_5\mathrm{C}_2 \cdot {}_6\mathrm{C}_2 \cdot {}_4\mathrm{C}_1 = 10 \cdot 15 \cdot 4 = 600\).
Step 3 — Add.
$$360 + 450 + 600 = 1{,}410.$$Answer: \(1{,}410\) samples.
Problem 18. Three of one kind and two of another kind.
Problems 7.6.19–7.6.24: Find the number of different ways to draw a 5-card hand from a deck to have the following combinations.
Problem 18 Solution
Step 1 — Pick the "three of" denomination and the "two of" denomination. The roles are distinguishable, so the six ordered pairs of distinct denominations are separate cases. Let \(D = 4, N = 5, P = 6\).
- 3D, 2N: \({}_4\mathrm{C}_3 \cdot {}_5\mathrm{C}_2 = 4 \cdot 10 = 40\). - 3D, 2P: \({}_4\mathrm{C}_3 \cdot {}_6\mathrm{C}_2 = 4 \cdot 15 = 60\). - 3N, 2D: \({}_5\mathrm{C}_3 \cdot {}_4\mathrm{C}_2 = 10 \cdot 6 = 60\). - 3N, 2P: \({}_5\mathrm{C}_3 \cdot {}_6\mathrm{C}_2 = 10 \cdot 15 = 150\). - 3P, 2D: \({}_6\mathrm{C}_3 \cdot {}_4\mathrm{C}_2 = 20 \cdot 6 = 120\). - 3P, 2N: \({}_6\mathrm{C}_3 \cdot {}_5\mathrm{C}_2 = 20 \cdot 10 = 200\).
Step 2 — Add.
$$40 + 60 + 60 + 150 + 120 + 200 = 630.$$Answer: \(630\) samples.
Problem 19. Three face cards.
Problem 19 Solution
Step 1 — There are 12 face cards (J, Q, K of each suit) and 40 non-face cards. Choose \(3\) face cards and \(2\) non-face cards.
Step 2 — Multiply the combinations.
$${}_{12}\mathrm{C}_3 \cdot {}_{40}\mathrm{C}_2 = 220 \cdot 780 = 171{,}600.$$Answer: \(171{,}600\) hands.
Problem 20. A heart flush (all hearts).
Problem 20 Solution
Step 1 — All 5 from the 13 hearts.
$${}_{13}\mathrm{C}_5 = 1{,}287.$$Answer: \(1{,}287\) hands.
Problem 21. Two hearts and three diamonds.
Problem 21 Solution
Step 1 — Two hearts from 13 and three diamonds from 13.
$${}_{13}\mathrm{C}_2 \cdot {}_{13}\mathrm{C}_3 = 78 \cdot 286 = 22{,}308.$$Answer: \(22{,}308\) hands.
Problem 22. Two cards of one suit, and three of another suit.
Problem 22 Solution
Step 1 — Choose the two suits (order matters). The "three-of-a-suit" role and the "two-of-a-suit" role are different, so we pick one suit for the three and a different suit for the two.
$${}_4\mathrm{C}_1 \cdot {}_3\mathrm{C}_1 = 4 \cdot 3 = 12 \text{ suit orderings.}$$Step 2 — Choose the cards within each suit.
$${}_{13}\mathrm{C}_3 \cdot {}_{13}\mathrm{C}_2 = 286 \cdot 78 = 22{,}308.$$Step 3 — Multiply.
$$12 \cdot 22{,}308 = 267{,}696.$$Answer: \(267{,}696\) hands.
Problem 23. Two kings and three queens.
Problem 23 Solution
Step 1 — Two kings from 4 and three queens from 4.
$${}_4\mathrm{C}_2 \cdot {}_4\mathrm{C}_3 = 6 \cdot 4 = 24.$$Answer: \(24\) hands.
Problem 24. 2 cards of one value and 3 of another value.
Problems 7.6.25–7.6.26: The party affiliation of the 100 United States Senators in the \(114^{\text{th}}\) Congress, January 2015, was: 44 Democrats, 54 Republicans, and 2 Independents.
Problem 24 Solution
Step 1 — Choose the two values (order matters). \(13\) choices for the value that appears three times, then \(12\) remaining choices for the value that appears twice.
Step 2 — Choose the cards within each value. Each value has \(4\) cards (one per suit).
$$13 \cdot {}_4\mathrm{C}_3 \cdot 12 \cdot {}_4\mathrm{C}_2 = 13 \cdot 4 \cdot 12 \cdot 6 = 3{,}744.$$Answer: \(3{,}744\) hands.
Problem 25. In how many ways could a 10-person committee be selected if it is to contain 4 Democrats, 5 Republicans, and 1 Independent?
Problem 25 Solution
Step 1 — One combination per party. Pick \(4\) of \(44\) Democrats, \(5\) of \(54\) Republicans, \(1\) of \(2\) Independents.
Step 2 — Evaluate each combination.
$${}_{44}\mathrm{C}_4 = 135{,}751, \quad {}_{54}\mathrm{C}_5 = 3{,}162{,}510, \quad {}_2\mathrm{C}_1 = 2.$$Step 3 — Multiply.
$$135{,}751 \cdot 3{,}162{,}510 \cdot 2 = 858{,}627{,}790{,}020.$$Answer: \(858{,}627{,}790{,}020\) committees.
Problem 26. In how many different ways could a 10-person committee be selected with 6 or 7 Republicans and the rest Democrats (with no Independents)?
Problems 7.6.27–7.6.28: The 100 United States Senators in the \(114^{\text{th}}\) Congress, January 2015, included 80 men and 20 women. Suppose a committee of senators is working on legislation about wage discrimination by gender.
Problem 26 Solution
Step 1 — Split on the number of Republicans. Case A: \(6\) Republicans and \(4\) Democrats. Case B: \(7\) Republicans and \(3\) Democrats.
Step 2 — Count each case.
$$\text{Case A: } {}_{54}\mathrm{C}_6 \cdot {}_{44}\mathrm{C}_4 = 25{,}827{,}165 \cdot 135{,}751 = 3{,}506{,}063{,}475{,}915.$$ $$\text{Case B: } {}_{54}\mathrm{C}_7 \cdot {}_{44}\mathrm{C}_3 = 177{,}100{,}560 \cdot 13{,}244 = 2{,}345{,}519{,}816{,}640.$$Step 3 — Add.
$$3{,}506{,}063{,}475{,}915 + 2{,}345{,}519{,}816{,}640 = 5{,}851{,}583{,}292{,}555.$$Answer: \(5{,}851{,}583{,}292{,}555\) committees.
Problem 27. In how many ways could a 12-person committee be selected to contain equal numbers of men and women?
Problem 27 Solution
Step 1 — Equal numbers means 6 men and 6 women.
$${}_{80}\mathrm{C}_6 \cdot {}_{20}\mathrm{C}_6 = 300{,}500{,}200 \cdot 38{,}760 = 11{,}647{,}387{,}752{,}000.$$Answer: \(11{,}647{,}387{,}752{,}000\) committees.
Problem 28. In how many ways could a 6-person committee be selected to contain fewer women than men?
Problems 7.6.29–7.6.32: Jorge has 6 rock songs, 7 rap songs and 4 country songs that he likes to listen to while he exercises. He randomly selects six (6) of these songs to create a playlist to listen to today while he exercises. How many different playlists of 6 songs can be selected that satisfy each of the following? (We care which songs are selected to be on the playlist, but not what order they are selected or listed in.)
Problem 28 Solution
Step 1 — Enumerate "fewer women than men" cases for a 6-person committee. Fewer women than men in a 6-person committee means the women count \(w\) satisfies \(w < 6 - w\), so \(w \in \{0, 1, 2\}\) and men count \(m = 6 - w\).
Step 2 — Count each case.
- \(w = 0, m = 6\): \({}_{80}\mathrm{C}_6 \cdot {}_{20}\mathrm{C}_0 = 300{,}500{,}200 \cdot 1 = 300{,}500{,}200\). - \(w = 1, m = 5\): \({}_{80}\mathrm{C}_5 \cdot {}_{20}\mathrm{C}_1 = 24{,}040{,}016 \cdot 20 = 480{,}800{,}320\). - \(w = 2, m = 4\): \({}_{80}\mathrm{C}_4 \cdot {}_{20}\mathrm{C}_2 = 1{,}581{,}580 \cdot 190 = 300{,}500{,}200\).
Step 3 — Add the disjoint cases.
$$300{,}500{,}200 + 480{,}800{,}320 + 300{,}500{,}200 = 1{,}081{,}800{,}720.$$Answer: \(1{,}081{,}800{,}720\) committees.
Problem 29. Playlist has 2 songs of each type.
Problem 29 Solution
Step 1 — Two of each type. \(6\) rock, \(7\) rap, \(4\) country.
$${}_6\mathrm{C}_2 \cdot {}_7\mathrm{C}_2 \cdot {}_4\mathrm{C}_2 = 15 \cdot 21 \cdot 6 = 1{,}890.$$Answer: \(1{,}890\) playlists.
Problem 30. Playlist has no country songs.
Problem 30 Solution
Step 1 — "No country songs" = choose all 6 from the 13 rock-and-rap songs.
$${}_{13}\mathrm{C}_6 = 1{,}716.$$Answer: \(1{,}716\) playlists.
Problem 31. Playlist has 3 rock, 2 rap, and 1 country song.
Problem 31 Solution
Step 1 — Pick 3 rock from 6, 2 rap from 7, 1 country from 4.
$${}_6\mathrm{C}_3 \cdot {}_7\mathrm{C}_2 \cdot {}_4\mathrm{C}_1 = 20 \cdot 21 \cdot 4 = 1{,}680.$$Answer: \(1{,}680\) playlists.
Problem 32. Playlist has 3 or 4 rock songs and all the rest are rap songs.
Problem 32 Solution
Step 1 — "3 or 4 rock, rest rap" splits into two disjoint cases. Case A: \(3\) rock and \(3\) rap. Case B: \(4\) rock and \(2\) rap.
Step 2 — Count each case.
$$\text{Case A: } {}_6\mathrm{C}_3 \cdot {}_7\mathrm{C}_3 = 20 \cdot 35 = 700.$$ $$\text{Case B: } {}_6\mathrm{C}_4 \cdot {}_7\mathrm{C}_2 = 15 \cdot 21 = 315.$$Step 3 — Add.
$$700 + 315 = 1{,}015.$$Answer: \(1{,}015\) playlists.