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Counting

Page history last edited by PBworks 13 years, 8 months ago

Learning Objectives

 

Problem 1 of 3

 

Charlene, Abriel and 6 of their friends attend a movie.

 

(a) In how may ways can they be seated in a row so that Charlene and Abriel do not sit next to each other?

 

(b) In how many of these ways are Charlene and Abriel sitting at either end of the row?

 

(c) After the movie they all went out to eat. They were seated at a round table. Again, Charlene and Abriel will not sit next to each other. In how many ways can everyone be seated at the table?

 

Solution

 

 

Problem 2 of 3

 

The 6th term of the binomial expansion of (x3 - 4/x)n contains x16.

 

(a) Find the value of n.

 

(b) How many terms are there in the full expansion?

 

(c) Does the expansion have a constant term? If so, which one is it?

 

Solution

 

 

Problem 3 of 3

 

Solve each of the following algebraically:

 

(a) (n - 1)! = 6(n - 3)!

 

(b) (n - 1)P2 = 72

 

(c) 35(n - 3)C3 = 4(n)C3

 

Solution

 

a)

 

(n-1)! = 6(n-3)

 

divide by (n-3)! have all n variables on oneside

(n-1)! / (n-3)! = 6(n-3)! / (n-3)!

 

the (n-3) cancel leaving (n-1)(n-2)

(n-1)(n-2)(n-3)! / (n-3)! = 6

 

multiply the binomials (n-1)(n-2)

n2 - 3n + 2 = 6

 

subtract 6 from both sides to make one side = 0

n2 - 3n - 4 = 0

 

factoring the equation

(n+1)(n-4) = 0

 

answers

n = -1 n = 4

 

knowing that n =/= -1, we discard it, there fore only n MUST = 4

 

 

b)

 

(n-1)P2 = 72

 

rewrite the equation in factorial form

(n-1)! / {(n-1) - 2}!

 

combining (n-1) with -2 into (n-3)

(n-1)! / (n-3)! = 72

 

the (n-3)! cancel leaving (n-1)(n-2)

(n-1)(n-2)(n-3)! / (n-3)! = 72

 

expanding the binomial (n-1)(n-2)

n2 - 2n - n + 2 = 72

 

combine n terms and subtract 72 from both sides

n2 - 3n - 70 = 0

 

factor the equation

(n+7)(n-10) = 0

 

answers

n = -7 n = 10

 

since n cannot be negative

n = 10

 

c)

 

35(n-3)C3 = 4nC3

 

rewrite the equation in factorial form

35{(n-3)! / (n-3-3)!3!} = 4{n! / (n-3)!3!

 

simplifying the knowns

35{(n-3)! / (n-6)!6} = 4{n! / (n-3)!6}

 

expand factorials and cancel common denominators

35{(n-3)(n-4)(n-5)(n-6)! / (n-6)!6 = 4{(n)(n-1)(n-2)(n-3)! / (n-3)!6}

 

multiply both sides by the reciprocals to cancel out the 6 denominator

(6 / (n-3)(n-4)(n-5)) 35(n-3)(n-4)(n-5) / 6 = 4(n)(n-1)(n-2) / 6 (6 / (n-3)(n-4)(n-5))

 

simplifying

35 = 4 {(n)(n-1)(n-2) / (n-3)(n-4)(n-5)}

 

expanding the n variables

35 = 4 {(n3 + n2 - n - 2n - 2n - n + 2) / (n3 + n2 - 3n + 12 - 3n + 15 - 5n - 4n + 20)}

 

 

simplifying and cancel common terms

35 = 4 {(n3 + n2 -6n + 2) / (n3 + n2 -15n + 47)}

 

isolate the variables

35 / 4 = 4 {(-6n + 2) / (-15n + 47)} / 4

 

 

cross multiply

-525n + 1645 = -24n + 8

 

bring all n's to one side and the rest to the other side

-501n = -1637

 

isolate n

n = -1637 / -501

 

solving

n = 3.2675 (approximiately)

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