factorial question (1 Viewer)

bouncing · Apr 14, 2010

so i know n!= n(n-1)(n-2)....3x2x1

so is (n-1)!=(n-1)(n-2)(n-3)...3x2x1
and (n-2)!=(n-2)(n-3)...3x2x1

how do you do this question :\ im getting a bit confused

nC2 + nC3 =220

i knwo itll be n!/2!(n-2)! + n/3!(n-3)!=220
can i get clarification on what the common denominator would be???
is it n-2 because n-3 falls into (n-2)!?

Drongoski · Apr 14, 2010

bouncing said:
so i know n!= n(n-1)(n-2)....3x2x1

so is (n-1)!=(n-1)(n-2)(n-3)...3x2x1
and (n-2)!=(n-2)(n-3)...3x2x1

how do you do this question :\ im getting a bit confused

nC2 + nC3 =220

i knwo itll be n!/2!(n-2)! + n/3!(n-3)!=220
can i get clarification on what the common denominator would be???
is it n-2 because n-3 falls into (n-2)!?

But what is the question? For nC2 + nC3 = 220, n must = 11

Pwnage101 · Apr 14, 2010

$^{n}C_{2}+^{n}C_{3}=220 \; \; \; \textup{[Such notation implies}\; n\; \textup{is a non-negative integer (strictly greater than 2), however such knowledge is not requred at HSC level]}$
$\textup{Thus:}\; \; \frac{n!}{2!\left ( n-2 \right )!}+\frac{n!}{3!\left ( n-3 \right )!}=\; \; \frac{n\left ( n-1 \right )}{2}+\frac{n\left ( n-1 \right )\left ( n-2 \right )}{6}=220$
$3n\left ( n-1 \right )+n\left ( n-1 \right )\left ( n-2 \right )=1,320$
$n\left ( n-1 \right )\left \{ 3+n-2 \right \}=1,320$
$n\left ( n-1 \right )\left $ n+1 \right $=1,320$
$\textup{This gives a polynomial of degree 3, so to approximate we will proceed by noting:}$
$\left ( n-1 \right )n\left ( n+1 \right )\approx n.n.n=n^{3}$
$\textup{Solving}\; \; n^{3}=1,320 \;\; \textup{gives}\; \; n=\sqrt[3]{1,320}\approx 10.9696\; \; \textup{which indicates our required answer could be }\; \; 11$
$\textup{Checking:}\; \; 10.11.12=1,320\; \; \textup{as required.}$
$\textup{Final check:}\; \; ^{11}C_{2}+^{11}C_{3}=55+165=220.$

CheesePlease · Apr 14, 2010

Beautiful solution Pwnage.

When you truly grasp "n!= n(n-1)(n-2)....3x2x1" rule and then realise n! = n(n-1)(n-2)(n-3)! factorials become a joy to work with. In this case it becomes simple cancelling down, when you have something like n!/(n-3)! it becomes simply n!(n-1)(n-2).

bouncing · Apr 15, 2010

Pwnage101 said:
$^{n}C_{2}+^{n}C_{3}=220 \; \; \; \textup{[Such notation implies}\; n\; \textup{is a non-negative integer (strictly greater than 2), however such knowledge is not requred at HSC level]}$
$\textup{Thus:}\; \; \frac{n!}{2!\left ( n-2 \right )!}+\frac{n!}{3!\left ( n-3 \right )!}=\; \; \frac{n\left ( n-1 \right )}{2}+\frac{n\left ( n-1 \right )\left ( n-2 \right )}{6}=220$
$3n\left ( n-1 \right )+n\left ( n-1 \right )\left ( n-2 \right )=1,320$
$n\left ( n-1 \right )\left \{ 3+n-2 \right \}=1,320$
$n\left ( n-1 \right )\left $ n+1 \right $=1,320$
$\textup{This gives a polynomial of degree 3, so to approximate we will proceed by noting:}$
$\left ( n-1 \right )n\left ( n+1 \right )\approx n.n.n=n^{3}$
$\textup{Solving}\; \; n^{3}=1,320 \;\; \textup{gives}\; \; n=\sqrt[3]{1,320}\approx 10.9696\; \; \textup{which indicates our required answer could be }\; \; 11$
$\textup{Checking:}\; \; 10.11.12=1,320\; \; \textup{as required.}$
$\textup{Final check:}\; \; ^{11}C_{2}+^{11}C_{3}=55+165=220.$

wow thank you, cant wait till that day i can use latex properly
i thought that n had to be strictly greater than 3 because it is nC3 (although im not too sure about anything right now since i've only just started this topic) thanks!

Pwnage101 · Apr 16, 2010

3C3 is valid, with nCr notation, n has to be a positive integer i.e. n=1,2,3,... and r has to be a non-negative integer up to (and can be including) n, i.e. r=0,1,2,3,...,n.

chrisnumber1 · Apr 16, 2010

bouncing said:
wow thank you, cant wait till that day i can use latex properly
i thought that n had to be strictly greater than 3 because it is nC3 (although im not too sure about anything right now since i've only just started this topic) thanks!

its strictly greater or equal to., as 3C3 is 1

factorial question (1 Viewer)

bouncing

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Drongoski

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CheesePlease

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bouncing

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