Answer to Question #236771 in Statistics and Probability for Lee3415

Question

Define/Illustrate Mutual Exclusivity and Independence.

3. Simplify fully:
(a) P(n+3 ,2), (2)
(b) C(2n ,2n-2) (2)

4. Solve for n: C(2n ,3)+P(n,1)=60 (5)

Answer 1

(a)

$$P(n+3, 2)=\dfrac{(n+3)!}{(n+3-2)!}=\dfrac{(n+3)!}{(n+1)!}$$

$$=(n+3)(n+2)=n^2+5n+15$$

(b)

$$C(2n, 2n-2)=\dfrac{(2n)!}{(2n-2)!(2n-(2n-2))!}$$

$$=\dfrac{2n(2n-1)}{1(2)}=n(2n-1)=2n^2-n$$

(c)

$$C(2n, 3)+P(n, 1)=60$$

$$\dfrac{(2n)!}{(3)!(2n-3)!}+\dfrac{n!}{(n-1)!}=60$$

$$2n(2n-1)(2n-2)+6n-360=0, n=2,3,4,...$$

$$4n^3-6n^2+2n+3n-180=0$$

$$4n^3-6n^2+5n-180=0$$

$$n=4:4(64)-6(16)+5(4)-180=0$$

$$0=0, True$$

$$4n^2(n-4)+10n(n-4)+45(n-4)=0$$

$$(n-4)(4n^2+10n+45)=0$$

Let $4n^2+10n+45=0$

$$D=(10)^2-4(4)(45)=-620<0$$

The equation has no real solutions.

Answer: $n=4.$