nCr / Biotheo question (1 Viewer)

[Starcraftmazter]

Sup, our 2nd lesson on binomial theorem today, and we were given a set to do, except I don't remember learning how to do these types of questions lol - or maybe I'm just so stupid, I don't realise I did? That is not the question....

Anyway, how do I go about doing for eg, these 2:

Q: Find n if
a) ⁸C₃ + ⁸C₄ = ⁿC₄
b) ⁿC₄ = ⁿC₅


Thanks 8)

 

[Riviet]

Use the fact that ⁿC_r = n!/{r!(n-r)!} and n!=n(n-1)(n-2)...3.2.1

 

[Starcraftmazter]

Thats what I tried to use originally, because I thought that's what we use for these types of questions, and then I got to this point (first question):

128 = [ n(n-1)(n-2)(n-3) ] / 24

Expanded looks like this:

128 = n⁴-6n³+11n²-6n

"And then what?", I asked myself, and thought there must be another way. But seeing as how you've said there is none, could you tell me what the next step is?


Thanks 8)

 

[Riviet]

(a)

LHS=8!/(5!3!) + 8!/(4!4!)

=4.8!/(5!4!) + 5.8!/(5!4!) [multiplying top and bottom of first fraction by 4 and multiplying top and bottom of second fraction by 5]

=9!/(5!4!)

=9!/{4!(9-4)!}

But RHS= n!/(4!(n-4)!)

.'. equating integers from LHS and RHS, n=9.

(b)

n!/{4!(n-4)!} = n!/{5!(n-5)!}

n! cancel out and note that (n-4)!=(n-4)(n-5)! so the next line becomes

1/4!(n-4) = 1/5!

n-4=5

n=9

 

[Starcraftmazter]

Yes yes yes - it all makes sense now!!!

Love ya man 8)

 

[pyrodude1031]

UMMZZ for part B .. there is actually a quicker way .. All you need to understand is the rule that nCr = (n-r)Cr . hence n=9 . thats bcos choosing 4 things from nine is the same as taking away 5 things from the four

 

[airie]

And use Pascal's triangle for part (a) too, seeing that a number in a row is equal to the two numbers directly above it in the previous row ie. ⁿC_r + ⁿC_r+1 = ⁿ⁺¹C_r+1. wtg for Pascal's triangle