Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

702 CHAPTER 12 Sequences and Series


Although it is possible to use Pascal’s triangle to find the coefficients in


for any positive integer value of n, it is impractical for large values of n. A more effi-


cient way to determine these coefficients uses the symbol n!(read“nfactorial”), de-


fined as follows.


1 x+ y 2 n


nFactorial (n!)

For any positive integer n,


By definition, 0!1.


n!n 1 n 121 n 221 n 32 Á 122112.


Evaluating Factorials

Evaluate each factorial.


(a) (b)


(c) 0! is defined to be 1.


FIGURE 3shows how a graphing calculator computes factorials. NOW TRY


Evaluating Expressions Involving Factorials

Find the value of each expression.


(a)


(b)


(c)


(d) NOW TRY


Now look again at the coefficients of the expansion


The coefficient of the second term is 5, and the exponents on the variables in that


term are 4 and 1. From Example 2(a), The coefficient of the third term is


10, and the exponents are 3 and 2. From Example 2(b), Similar results are


true for the remaining terms. The first term can be written as and the last term


can be written as Then the coefficient of the first term should be and


the coefficient of the last term would be


The coefficient of a term in in which the variable part is is


This is called a binomial coefficient.

The binomial coefficient is often represented by the symbol This nota-


tion comes from the fact that if we choose combinationsof nthings taken rat a time,


the result is given by that expression. We read as “combinations ofnthings


taken rat a time.”Another common representation is A.


n

rB


nCr


nCr^.


n!
r! 1 n-r 2!

n!


r! 1 n- r 2!


.


1 x+y 2 n xryn-r


5!

0!5!= 1.


5!

1 x 5!0!=1,


(^0) y (^5).


1 x^5 y^0 ,


5!

3!2!=10.


5!

4!1!=5.


1 x+y 25 = x^5 + 5 x^4 y+ 10 x^3 y^2 + 10 x^2 y^3 + 5 xy^4 + y^5.


4!


4!0!


=


4 # 3 # 2 # 1
14 # 3 # 2 # 12112

= 1


6!


3!3!


=


6 # 5 # 4 # 3 # 2 # 1
13 # 2 # 1213 # 2 # 12

=


6 # 5 # 4
3 # 2 # 1

= 20


5!


3!2!


=


5 # 4 # 3 # 2 # 1
13 # 2 # 1212 # 12

=


5 # 4
2 # 1

= 10


5!


4!1!


=


5 # 4 # 3 # 2 # 1
14 # 3 # 2 # 12112

= 5


EXAMPLE 2


0!= 1


3!= 3 # 2 # 1 = 6 5!= 5 # 4 # 3 # 2 # 1 = 120


EXAMPLE 1


FIGURE 3

NOW TRY
EXERCISE 1
Evaluate.
7!


NOW TRY ANSWERS



  1. 5040

  2. (a) 28 (b) 56 (c) 1 (d) 6


NOW TRY
EXERCISE 2
Find the value of each
expression.


(a) (b)


(c) (d)


6!


5!1!


6!


6!0!


8!


5!3!


8!


6!2!

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