04.20.10 Combinations.notebook
OBJECTIVE
Students will be able to determine the number of possible combinations for a given situation.
Students will be able to apply the Binomial Theorem.
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April 20, 2010
Answers to Homework page 705
16) 15
18) 1512
20) a) 45,697,600
b) 32,292,000
22) a) 118,813,760
b) 78,936,000
24) 120
26) 362,880
28) 5040
30) 479,001,600
32) 20
34) 5040
36) 3024
38) 1
40) 6
42) 120
44) 5040
46) 362,880
48) 20
50) 180
52) 210
54) 1,663,200
56) 2772
Combination
The number of combinations of r objects taken from a group of n
distinct objects is denoted by nCr.
nCr = n!
(n ­ r)! r!
Order does not matter
Combinations vs Permutations
The school board has seven members.
The board must have three officers: a chairperson, an assistant chairperson, and a secretary. 7C2
=
7!
(7 - 2)! 2!
7C2
= 21
4 C2
=
4!
(4 - 2)! 2!
4C2
= 6
(a) How many different sets of these officers can be formed from this board?
(b) How many three­person committees can be formed from this board?
Is part (a) asking for a number of permutaions or a number of combinations? What about part (b)?
1
04.20.10 Combinations.notebook
Binomial Theorem
(a + b)n means
Properties of the Binomial Expansion (a + b)n
• There are n + 1 terms. n
Ʃ
April 20, 2010
nCr
an-r br
r=0
• The first term is an and the final term is bn. • Progressing from the first term to the last, the exponent of a decreases by 1 from term to term while the exponent of b increases by 1. In addition, the sum of the exponents of a and b in each term is n. • If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term. 2
Expand (x
+ 3)6 (x2 + 3)6 = 6C0 (x2)6(3)0 + 6C1 (x2)5(3)1 + 6C2 (x2)4(3)2 + 6C3 (x2)3(3)3 + 6C4 (x2)2(3)4 + 6C5 (x2)1(3)5 + 6C6 (x2)0(3)6
(1)(x12)(1) + (6)(x10)(3) + (15)(x8)(9) + (20)(x6)(27) + (15)(x4)(81) + (6)(x2)(243) + (1)(1)(729)
= x12 + 18x10 + 135x8 + 540x6 + 1215x4 + 1458x2 + 729
2
04.20.10 Combinations.notebook
April 20, 2010
7
Expand (2x – 5y)
(2x – 5y)7 = 7C0 (2x)7(–5y)0 + 7C1 (2x)6(–5y)1 + 7C2 (2x)5(–5y)2 + 7C3 (2x)4(–5y)3 + 7C4 (2x)3(–5y)4 + 7C5 (2x)2(–5y)5 + 7C6 (2x)1(–5y)6 + 7C7 (2x)0(–5y)7
Then simplifying gives me: Copyright © Elizabeth Stapel 1999­2009 All Rights Reserved
(1)(128x7)(1) + (7)(64x6)(–5y) + (21)(32x5)(25y2) + (35)(16x4)(–125y3) + (35)(8x3)(625y4) + (21)(4x2)(–3125y5) + (7)(2x)(15625y6) + (1)(1)(–78125y7)
= 128x7 – 2240x6y + 16800x5y2 – 70000x4y3 + 175000x3y4 – 262500x2y5 + 218750xy6 – 78125y7
10
What is the fourth term in the expansion of (3x – 2) ?
(3x – 2)10 = 10C0 (3x)10–0(–2)0 + 10C1 (3x)10–1(–2)1 + 10C2 (3x)10–2(–2)2 + 10C3 (3x)10–3(–2)3 + 10C4 (3x)10–4(–2)4 + 10C5 (3x)10–5(–2)5 + 10C6 (3x)10–6(–2)6 + 10C7 (3x)10–7(–2)7 + 10C8 (3x)10–8(–2)8 + 10C9 (3x)10–9(–2)9 + 10C10 (3x)10–10(–2)10 12
Find the tenth term in the expansion of (x + 3) . To find the tenth term, I plug x, 3, and 12 into the Binomial Theorem, using the number 10 – 1 = 9 as my counter:
12C9 (x)
12–9
(3)9 = (220)x3(19683) = 4330260x3
Note that, in any expansion, there is one more term than the number in the power. For instance:
(x + y)2 = x2 + 2xy + y2 (second power: three terms)
(x + y)3 = x3 + 3x2y + 3xy2 + y3 (third power: four terms)
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 (fourth power: five terms)
The expansion in this exercise, (3x – 2)10, has power of n = 10, so the expansion will have eleven terms, and the terms will count up, not from 1 to 10 or from 1 to 11, but from 0 to 10. This is why the fourth term will not the one where I'm using "4" as my counter, but will be the one where I'm using "3".
10–3
(–2)3 = (120)(2187)(x7)(–8) = –2099520x7
10C3 (3x)
8
Find the middle term in the expansion of (4x – y) .
Since this binomial is to the power 8, there will be nine terms in the expansion, which makes the fifth term the middle one. So I'll plug 4x, –y, and 8 into the Binomial Theorem, using the number 5 – 1 = 4 as my counter.
C4 (4x)8–4(–y)4 = (70)(256x4)(y4) = 17920x4y4
8
HOMEWORK
Chapter 12 - Lesson 2
page 712
# 18 - 46 even
(15 problems)
3