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 6^{th} term of the binomial expansion of (*x*^{3} - 4/*x*)^{n} contains *x*^{16}.

(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)}P_{2} = 72

(c) 35_{(n - 3)}C_{3} = 4_{(n)}C_{3}

**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)**

n^{2} - 3n + 2 = 6

**subtract 6 from both sides to make one side = 0**

n^{2} - 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)}P_{2} = 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)**

n^{2} - 2n - n + 2 = 72

**combine n terms and subtract 72 from both sides**

n^{2} - 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)}C_{3} = 4_{n}C_{3}

**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 {(n^{3} + n^{2} - n - 2n - 2n - n + 2) / (n^{3} + n^{2} - 3n + 12 - 3n + 15 - 5n - 4n + 20)}

**simplifying and cancel common terms**

35 = 4 {(~~n~~^{3} + ~~n~~^{2} -6n + 2) / (~~n~~^{3} + ~~n~~^{2} -15n + 47)}

**isolate the variables**

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

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