2 QR Problems...can you solve them?

[Eddie830]

Does anyone know how to solve these two:

1) Chords AB and CD of Circle O intersect at E. If AE=4, AB=5, CE=2, find ED.

2) A florist purchased 3 yellow roses, 2 pink roses, and 5 red roses. How many 3 rose arrangements are possible?

I have asked a few ppl already and no one can figure these two out...

Answers below:









1) 2

2) 120

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[HBomb]

To answer 1) use this: when two chords intersect, the lengths of the chord sections always multiply to the same value. For this problem AE x EB = CE x ED. That's the only hard part, to know that this exists. Then the problem is easy to solve. Since AE is 4 and AB is 5, then EB is 1. Since CE is 2, you get (4)(1) = 2(ED), so ED = 2.

I remember this problem when I was studying for the DAT, and I didn't know how to do it at the time. But then I just typed up some of the keywords into google and presto, found the answer in minutes. Just wanted to say this because I think it's good advice...to use the internet because it can be a powerful tool.

I don't know about #2. Maybe someone else knows. At least let me think about it for a while.

 

[time4teeth]

how can 120 be the answer for the flower arrangements? obviously order doesn't matter because you don't put flowers in a row, you bunch them.

Also, who a yellow flower is a yellow flower--- doesn't matter which one.

so with 3 yellow, 2 pink, 5 red:

yyy
ypr
rrr
ypp
yrr
pyy
prr
ryy
rpp

am is totally wrong on this?

lemme know......

 

[Slain]

how can 120 be the answer for the flower arrangements? obviously order doesn't matter because you don't put flowers in a row, you bunch them.

Also, who a yellow flower is a yellow flower--- doesn't matter which one.

so with 3 yellow, 2 pink, 5 red:

yyy
ypr
rrr
ypp
yrr
pyy
prr
ryy
rpp

am is totally wrong on this?

lemme know......

Yeah it is wrong. The questions asks for how many ROSE combinations you can have, not how many "colored rose" combinations you can have.

You have 10 roses so:

10!/3!(10-3)! = 120 ----- This is of course a continuation of the following equation that you must know for the DAT

x!/Y!(x-y)! where x > y

 

[HBomb]

all this time, it was just a straight up combinations problem!!! who would've guessed?

 

[Eddie830]

seriously!

 

[DrTacoElf]

Topscore exam 2 i see

 

[drusier]

thanks for explaining that rose problem however, i was wondering if the equation can be simplified so on test day i dont have to work everthing out?
x!/Y!(x-y)! where x > y??

 

[bpenly]

Drusier

Here is how you simplify:

you have 10!/(10-3)! * 3!

you can simplify so that it is 10!/(7!)*(3!)

You can pull a cute trick here when you basically "subtract" a denominator from the top and eliminate it from the bottom. Picking 7! you get:

10 x 9 x 8/3!


= 720/6
= 120 (answer)

I know that my terminology is bad, but this works. That is the important thing.

 

[boogaking]

Combinatorial rule....

X!/Y!

X= number of things involved eg number of roses

Y= number of different positions eg each rose can be in only one position of that trio of roses.... arrangement of three

so 10!/ 3! = 10 x 9x 8/ 3x2x1 => 120!!!!!

 

[HBomb]

Combinatorial rule....

X!/Y!

X= number of things involved eg number of roses

Y= number of different positions eg each rose can be in only one position of that trio of roses.... arrangement of three

so 10!/ 3! = 10 x 9x 8/ 3x2x1 => 120!!!!!

Sorry, what you said isn't right (you likely made a typo):
Combination (where order doesn't matter):
C(X,Y) = X!/[Y!(X-Y)!]
so
C(10,3) = 10!/[3!(10-3)!] = 10!/[3!7!] (and then as you state equals 10 x 9x 8/ 3x2x1 => 120)

Also:
Permutaion (where order matters):
P(X,Y) = X!/(X-Y)!

 

[x-linked]

i dont understand why it is 120, i thought it was 10. coz all 3 yellow flowers are the same. not three different things so i dont think 10C3 can apply here.