Cardmeister: Solutions

Combination Practice Problems

Problems with Solutions

1. At Louie's Licks, the following ice cream flavors and cones are offered:

Flavors	Cones
Chocolate	Waffle
Vanilla	Sugar
Mint Chocolate Chip	Cake
Strawberry
Rocky Road

For $1.50, a customer may get "Louie's Delight" - the cone of his or her choice and with two different scoops of ice cream. How many different ways may a person order "Louie's Delight" ?

For $2.50, a customer may get "Monster Madness" - the cone of his or her choice with 3 scoops of ice cream. With this special, one may order the same ice cream more than once. How many different ways can a person order "Monster Madness"?

Solution

There are

(	5	)	= 10
	2

ways to choose 2 flavors from a total of 5. Since one may have these flavors for each of 3 cone choices, there are 3 * 10 = 30 different ways to order Louie's Delight.

With Monster Madness, one may choose 3 scoops of ice cream; in addition, one may choose the same ice cream more than once. Thus there are 5 * 3 = 15 total options for ice cream. Since one is permitted to choose

(	15	)	= 286
	3

different choices for 3 scoops of ice cream. Since there are 286 ice cream choices for 3 different cone choices, there are 286 * 3 = 858 ways to order Monster Madness.

2. Mr. Pascal must select an ice hockey team of 9 players from 17 seniors and 15 juniors. How many ways may he select a team if...

he has no restrictions?
two seniors, Anthony and Marc-Eric are the best goalies and must be on the team?
he must pick 5 seniors and 4 juniors?

There are 17 + 15 = 32 players to choose from, and Mr. Pascal must select 9 players for the team. With no restrictions, there are

(	32	)	= 28,048,800
	9

= 28,048,800 different teams possible.

If two seniors must be on the team, then only 7 more players may be picked. There are 15 seniors and 15 juniors which may be chosen for the remaining 7 slots, so there are

(	30	)	= 2,035,800
	7

possible teams with Anthony and Marc-Eric.

If 5 seniors must be picked, then there are

(	17	)	= 6188
	5

ways to pick the seniors. For each of these selections, there are

(	15	)	= 1365
	4

ways to select four juniors. Thus there are 6188 * 1365 = 8,446,620 different line-ups with 5 seniors and 4 juniors.

3. Shui is taking her biology quiz, and is instructed to answer 7 out of the 10 questions. How many different ways may she answer the questions if...

she has no restrictions?
she must answer 4 out of the first 5, and 3 out of the last 5?
she must answer at least 3 of the first 5?

If Shui has no restrictions, she may answer 7 out of 10 questions in

(	10	)	= 120
	7

different ways.

There are

(	5	)	(	5	)	= 50
	4			3

ways to answer the test. There are C(5,4) = 5 ways to answer four of the first five questions; for each of these ways to answer the first half of the test, there are C(5,3) = 10 ways to answer 3 out of the last 5 questions. Thus there are 5 * 10 = 50 ways to complete the exam.

If at least three of the first five questions must be answered, there are three cases to consider: answering 3 out of the first five, answering 4 out of the first five, and answering all of the first five.

Answering 3 of the first five implies that Shui must answer 4 of the second five. She may do this in

(	5	)	(	5	)	= 50
	3			4

ways to answer 3 of the first 5. Similarly, there are

(	5	)	(	5	)	= 50
	4			3

ways to choose four out of the first five and

(	5	)	(	5	)	= 10
	5			2

ways to choose all five of the first five. Adding up all three, we obtain 50 + 50 + 10 = 110 ways to answer at least 3 of the first five questions on the examination.

4. The English alphabet has 26 letters - five of those are vowels, the others are consonants. For this question, consider a "word" to have no two letters which are the same - thus "funny" would not be a "word" since it has two n's.

How many 5 letter "words" are there?
What is the number of 10 letter "words" that have 4 or more vowels?
How many 7 letter "words" start with 'x' and have either an 'e' or an 'i' in them?

There are 26 letters total from which one must choose 5, thus there are

(	26	)	= 65,780
	5

combinations possible for "words" with five letters.

To find the number of 10 letter "words" with 4 or more vowels, we must count the number of 10 letter words that have exactly 4 vowels, and those that have exactly five vowels. (There may be no more than five vowels in the word, or else one of the vowels would be repeated). Thus there are

(	21	)	(	5	)	= 271,320
	6			4

"words" 10 letter words with four vowels. There are

(	21	)	(	5	)	= 20,349
	5			5

different 10 letter "words" with all five vowels. Adding the number of 10 letter "words" with (exactly) four vowels to the number of 10 letter "words" with five vowels gives 271,320 + 20,349 = 291,669 "words" that have at least four vowels.

To find the total number of 7 letter words starting with 'x', find the number of 6 letter words not containing 'x', and then place and 'x' in front of each of those words. Thus, there are C(25,6) = 177,100 words that begin with 'x.'

To find the number of words that have an 'x' and an 'e', find the number of five letter words that do not have 'x' or 'e', and then place 'xe' in front of those words. There are C(24,5) 7 letter words, then, which have 'xe' in them. Similarly, there are also C(24,5) words that have 'xi' in them. However, we cannot simply add C(24,5) + C(24,5) to get the number of words that contain 'xe' or 'xi'. Consider the word "xerion." It has both an 'e' and an 'i' in it - so it was counted twice. Thus we must subtract out all combinations of words that contain 'xei.' There are C(23, 4) of these seven letter words. Thus, there are

C(24,5) + C(24,5) - C(23, 4) = 76,153 words

that have either 'xe' or 'xi' in them.

5. Jacie is playing a game of poker, where she is dealt 5 cards from the usual deck of 52. How many ways may she be dealt (a) four of a kind? (b) a pair of jacks?

Solution

To have a four of a kind, Jacie must select 4 of the same card (ie four 8's) and then from the remaining 52 - 4 = 48 cards, she must choose one card left. There are 13 ways to choose four of a kind: four 2's, four 3's, ... four Queens, four Kings, four Aces. For each of these 13 ways, there are 48 ways to choose one of the four cards that are remaining. Thus there are 13 * 48 = 624 ways to be dealt four of a kind.

There are C(4,2) = 6 ways to choose a pair of Jacks:

For each of these pairs, there are C(50, 3) = 19,600 ways to receive the other 3 cards from the remaining 50. Thus there are 6 * 19,600 = 117,600 ways to receive a pair of Jacks.

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