1. Evaluate the expression.
1. 5!
ANSWER
120
2. (4 – 2) !3!
ANSWER 12

2. Evaluate the expression.5! 8! 3.
ANSWER
336
4. 7 P5
ANSWER
2520

3. Evaluate the expression.
5. In how many ways can 6 people line up to buy tickets for a movie?
ANSWER
720

4. EXAMPLE 1
Find combinations
Cards
A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit.
If the order in which the cards are dealt is not important, how many different 5-card hands are possible?
In how many 5-card hands are all 5 cards of the same color?

5. EXAMPLE 1
Find combinations
SOLUTION
The number of ways to choose 5 cards from a deck of 52 cards is:
47! 5! =
52!
52C5 = !
47! 5!
= 2,598,960.

6. EXAMPLE 1
Find combinations
For all 5 cards to be the same color, you need to choose 1 of the 2 colors and then 5 of the 26 cards in that color. So, the number of possible hands is:
21! 5! =
26!
26C5
2C1
1! ! 2! = !
21! 5!
1! ! 2
= 131,560.

7. EXAMPLE 2
Decide to multiply or add combinations
Theater
William Shakespeare wrote 38 plays that can be divided into three genres. Of the 38 plays, 18 are comedies, 10 are histories, and 10 are tragedies.
How many different sets of exactly 2 comedies and 1 tragedy can you read?
How many different sets of at most 3 plays can you read?

8. EXAMPLE 2
Decide to multiply or add combinations
SOLUTION
You can choose 2 of the 18 comedies and 1 of the 10 tragedies. So, the number of possible sets of plays is:
9! 1! =
10!
10C1
18C2
16! !
18! = !
16! 9! ! = =

9. EXAMPLE 2
Decide to multiply or add combinations
You can read 0, 1, 2, or 3 plays. Because there are 38 plays that can be chosen, the number of possible sets of plays is:
38C0 + 38C1 + 38C2 +38C3 = =

10. EXAMPLE 3
Solve a multi-step problem
Basketball
During the school year, the girl’s basketball team is scheduled to play 12 home games. You want to attend at least 3 of the games. How many different combinations of games can you attend?
SOLUTION
Of the 12 home games, you want to attend 3 games, or 4 games, or 5 games, and so on. So, the number of combinations of games you can attend is:
12C3 + 12C4 + 12C5 +………..+ 12C12

11. EXAMPLE 3
Solve a multi-step problem
Instead of adding these combinations, use the following reasoning. For each of the 12 games, you can choose to attend or not attend the game, so there are 212 total combinations. If you attend at least 3 games, you do not attend only a total of 0, 1, or 2 games. So, the number of ways you can attend at least 3 games is:
212 –
(12C0 + 12C1 + 12C2 )
= – ( ) =

12. YOU TRY
for Examples 1, 2 and 3
Find the number of combinations.
8C3 1.
SOLUTION
(8 – 3)! 3! = 8!
8C3 = !
5! 3! 6 =
336 = 56

13. YOU TRY
for Examples 1, 2 and 3
Find the number of combinations. 2.
10C6
SOLUTION
(10 – 6)! 6! =
10!
10C6 = !
4! 6! =
5040 24 =
210

14. YOU TRY
for Examples 1, 2 and 3
Find the number of combinations.
7C2 3.
SOLUTION =
7C2
(7 – 2)! 2! 7! = !
5! 2! = 42 2 = 21

15. YOU TRY
for Examples 1, 2 and 3
Find the number of combinations. 4.
14C5
SOLUTION
(14 – 5)! 5! =
14!
14C5
9! 5! = ! =
2002

16. YOU TRY
for Examples 1, 2 and 3
What If? In Example 2, how many different sets of exactly 3 tragedies and 2 histories can you read?
SOLUTION
10C2
10C3
(10 – 3)! 3! =
10!
(10 – 2)! 2!
8! 2! =
10!
7! ! = !
7! 3!
8! 2! ! =
720 6 90 2 =
= sets

20. WARM UP
Find 9C5. 1.
ANSWER
126 2.
The manager of a chain of restaurants must choose 6 restaurants from 11 for a promotion. How many different selection can be made?
ANSWER
462 selections

21. WARM UP 3.
A committee consists of 10 Republicans and 8 Democrats. In how many ways can a subcommittee be chosen if it has 5 Republicans and 4 Democrats?
ANSWER
17,640 ways

22. EXAMPLE 4
Use Pascal’s triangle
School Clubs
The 6 members of a Model UN club must choose 2 representatives to attend a state convention. Use Pascal’s triangle to find the number of combinations of 2 members that can be chosen as representatives.
SOLUTION
Because you need to find 6C2, write the 6th row of Pascal’s triangle by adding numbers from the previous row.

23. EXAMPLE 4
Use Pascal’s triangle
n = 5 (5th row)
n = 6 (6th row)
6C0
6C1
6C2
6C3
6C4
6C5
6C6
ANSWER
The value of 6C2 is the third number in the 6th row of Pascal’s triangle, as shown above. Therefore, 6C2 = 15. There are 15 combinations of representatives for the convention.

24. YOU TRY
for Example 4 6.
What If? In Example 4, use Pascal’s triangle to find the number of combinations of 2 members that can be chosen if the Model UN club has 7 members.
SOLUTION
Because you need to find 7C2, write the 7th row of Pascal’s triangle by adding numbers from the previous row.
n = 6 (6th row)
n = 7 (7th row)
7C0
7C1
7C2
7C3
7C4
7C5
7C6
7C7

25. YOU TRY
for Example 4
ANSWER
The value of 7C2 is the third number in the 7th row of Pascal’s triangle, as shown above. Therefore, 7C2 = 21. There are 21 combinations of representatives for the convention.

26. EXAMPLE 5
Expand a power of a binomial sum
Use the binomial theorem to write the binomial expansion.
(x2 + y)3
= 3C0(x2)3y0 + 3C1(x2)2y1 + 3C2(x2)1y2 + 3C3(x2)0y3
= (1)(x6)(1) + (3)(x4)(y) + (3)(x2)(y2) + (1)(1)(y3)
= x6 + 3x4y + 3x2y2 + y3

27. EXAMPLE 6
Expand a power of a binomial difference
Use the binomial theorem to write the binomial expansion.
(a –2b)4
= [a + (– 2b)]4
= 4C0a4(–2b)0 + 4C1a3(–2b)1 + 4C2a2(–2b)2 + 4C3a1(–2b)3
+ 4C4a0(–2b)4
= (1)(a4)(1) + (4)(a3)(–2b) + (6)(a2)(4b2) + (4)(a)(–8b3)
+ (1)(1)(16b4)
= a4 – 8a3b + 24a2b2 – 32ab3 + 16b4

28. YOU TRY
for Examples 5 and 6
Use the binomial theorem to write the binomial expansion.
(x + 3)5
SOLUTION
For any positive integer n,the binomial expansion of (a + b)n is
(a + b)n =
nC0 anb0 + nC1 an–1b1 + nC2 an–2b2 + ….. +nCn a0bn.
(x + 3)5
5C0(x5)(30) + 5C1(x5–1)(31) + 5C2 (x5–2)(32) + =
5C3 (x5–3)(33) + 5C4 (x5–4)(34) + 5C5 (x5–5)(35)

29. YOU TRY
for Examples 5 and 6
(1)(x5)(1) + (5)(x4)(3) + (10)(x3)(9) + (10)(x2)(270) + =
5(x1)(81) + 1(x0)(243)
x5 + 15x4+ 90x x x + 243 =

30. YOU TRY
for Examples 5 and 6
Use the binomial theorem to write the binomial expansion.
(a + 2b)4
SOLUTION
For any positive integer n,the binomial expansion of (a + b)n is
(a + b)n =
nC0 anb0 + nC1 an–1b1 + nC2 an–2b2 + ….. +nCn a0bn. =
(a + 2b)4
4C0a4(2b)0 + 4C1a3(2b)1 + 4C2a2(2b)2 + 4C3a1(2b)3
+ 4C4a0(2b)4

31. YOU TRY
for Examples 5 and 6 =
(1)(a4)(1) + (4)(a3)(2b) + (6)(a2)(4b2) + (4)(a)(8b3)
+ (1)(1)(16b4) =
a4 + 8a3b + 24a2b2 + 32ab3 + 16b4

32. GUIDED PRACTICE
for Examples 5 and 6
Use the binomial theorem to write the binomial expansion.
(2p – q)4
SOLUTION
For any positive integer n,the binomial expansion of (a + b)n is
(a + b)n =
nC0 anb0 + nC1 an–1b1 + nC2 an–2b2 + ….. +nCn a0bn. =
(2p – q)4
[2p + (– q)]4
4C3 (2p)1(–q)3 + 4C4 (2p) 0(–q)4
4C0 (2p)4(–q)0 + 4C1 (2p)3(–q)1 + 4C2 (2p)2(–q)2 + =

33. YOU TRY
for Examples 5 and 6 =
(1)(16p4)(1) + (4)(8p3)(–q) + (6)(4p2)(q2) +
(16)(2p)(–q3) + (1)(1)(q4) =
16p4 – 32p3q + 24p2q2 – 8pq3 + q4

34. YOU TRY
for Examples 5 and 6
Use the binomial theorem to write the binomial expansion.
(5 – 2y)3
SOLUTION
For any positive integer n,the binomial expansion of (a + b)n is
(a + b)n =
nC0 anb0 + nC1 an–1b1 + nC2 an–2b2 + ….. +nCn a0bn. =
(5 – 2y)3
[5 + (– 2y)]3
3C0 (5)3(–2y)0 + 3C1 (5)2(–2y)1 + 3C2 (5)1(–2y)2
+ 3C3 (5)0(–2y)3 =

35. YOU TRY
for Examples 5 and 6 =
(1)(125)(1) + (3)(25)(–2y) + (6)(5)(4y2) + (1)(1)(–8y3) =
– 8y3 + 60y2 – 150y + 125

36. EXAMPLE 7
Find a coefficient in an expansion
Find the coefficient of x4 in the expansion of (3x + 2)10.
SOLUTION
From the binomial theorem, you know the following:
(3x + 2)10
= 10C0(3x)10(2)0 + 10C1(3x)9(2) C10(3x)0(2)10
Each term in the expansion has the form 10Cr(3x)10 – r (2) r. The term containing x4 occurs when r = 6:
10C6(3x)4(2)6
= (210)(81x4)(64)
= 1,088,640x4
ANSWER
The coefficient of x4 is 1,088,640.

37. YOU TRY
for Example 7
Find the coefficient of x5 in the expansion of (x – 3)7.
11.
SOLUTION
From the binomial theorem, you know the following:
(x – 3)7
7C0(x)7(–3)0 + 7C1(x)6(–3) C7(x)0(–3)7 =
Each term in the expansion has the form 7Cr(x)7 – r (–3)r. The term containing x5 occurs when r = 2:
7Cr(x)7 – r (–3)r
= (21)(x5)(9)
= 189x5
ANSWER
The coefficient of x5 is 189.

38. YOU TRY
for Example 7
12.
Find the coefficient of x3 in the expansion of (2x + 5)8.
SOLUTION
From the binomial theorem, you know the following: =
(2x + 5)8
8C0(2x)8(5)0 + 8C1(2x)7(5) C8(2x)0(5)8
Each term in the expansion has the form 8Cr(2x)8 – r (5)r. The term containing x3 occurs when r = 5:
8C5(2x)3(5)5
= (56)(8x3)(3125)
= 1,400,000x3
ANSWER
The coefficient of x3 is 1,400,000.