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Combination & Permutation

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Combination & Permutation

by chipjet » Mon Mar 22, 2010 10:20 am
Can someone please explain how to calculate this answer for me?

Suppose you have 15 accessories in a standard computer buildup and each accessory has 3 options (ex. Hard drive space could be 100GB, 200GB, or 300GB, RAM could be 1MB, 2MB or 3MB, monitor could be 15", 17" or none, etc.) How many different computers could be built?

Thanks in advance.
Last edited by chipjet on Mon Mar 22, 2010 11:05 am, edited 1 time in total.
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by analyst218 » Mon Mar 22, 2010 11:03 am
chipjet wrote:Can someone please explain how to calculate this answer for me?

Suppose you have 15 accessories in a standard computer buildup and each accessory has 3 options (ex. Hard drive space could be 100GB, 200GB, or 300GB, etc.) How many different computers could be built?

Thanks in advance.
the question is kind of vague, but assuming each buildup consists of single accesory,
15x3 = 45 possible buildups
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by chipjet » Mon Mar 22, 2010 11:08 am
It has to be more than that.

For example, just with 2 accessories with 3 options each, there are 6 possible buildups. With 3 accessories with 3 options each, there are 27 possible buildups.

I just don't know how to do the math to figure out when there are 15 accessories that all have to be accounted for on each buildup.
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by yeahdisk » Mon Mar 22, 2010 12:52 pm
When you have n things to choose from ... you have n choices each time!

So when choosing r of them, the permutations are:

n × n × ... (r times) = n^r

(Because there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on.)

3^15
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by analyst218 » Mon Mar 22, 2010 1:12 pm
chipjet wrote:It has to be more than that.

For example, just with 2 accessories with 3 options each, there are 6 possible buildups. With 3 accessories with 3 options each, there are 27 possible buildups.

I just don't know how to do the math to figure out when there are 15 accessories that all have to be accounted for on each buildup.
the question is poorly composed.
it does not state whether you can build a computer using any number of accessories(1~15)
how r u calculating 3 accessories = 27 buildups?

shouldnt it be 3(1+...+15) = 360
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by chipjet » Mon Mar 22, 2010 1:19 pm
analyst218 wrote:
chipjet wrote:It has to be more than that.

For example, just with 2 accessories with 3 options each, there are 6 possible buildups. With 3 accessories with 3 options each, there are 27 possible buildups.

I just don't know how to do the math to figure out when there are 15 accessories that all have to be accounted for on each buildup.
the question is poorly composed.
it does not state whether you can build a computer using any number of accessories(1~15)
how r u calculating 3 accessories = 27 buildups?

shouldnt it be 3(1+...+15) = 360
Yeah, sorry the question is a bit confusing. I didn't know a better way to build it up. I'll show you the 3 accessories with 3 options each = 27.

Accessories (F,G,H)
Options on F = Fa, Fb, Fc
Options on G = Ga, Gb, Gc
Options on H = Ha, Hb, Hc

So, to get the buildups, you would say:

Fa,Ga,Ha
Fa, Ga,Hb
Ha,Ga,Hc

Fa,Gb,Ha
Fa,Gb,Hb
Fa,Gb,Hc

Fa,Gc,Ha
Fa,Gc,Hb
Fa,Gc,Hc

....and so on... but I just couldn't remember the formula for calculating this with so many more accessories because instead of 3 accessories, there are 15 (with 3 options each).
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by Bha148 » Mon Mar 22, 2010 11:22 pm
1. As the question says standard computer is built using 15 different parts, we can assume that inorder to build a new computer we need 15 parts.
2. If all the 15 parts are having 3 types to choose from then it comes down to simple formula N^r
For example if we have 3 digit lock (all having numbers 0 to 9). total number of combinations possible are 10^3
So similarly for 15 parts computer, total combinations will be 15^3.
Knowledge is Power !
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by nysnowboard » Mon Mar 22, 2010 11:59 pm
yeahdisk wrote:When you have n things to choose from ... you have n choices each time!

So when choosing r of them, the permutations are:

n × n × ... (r times) = n^r

(Because there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on.)

3^15
If I may interject, although I am no expert, after thinking about it for a couple minutes I think yeahdisk is right...

Draw out a decision tree (or whatever it's called) starting with Accessory 1. From that starting node you have three possible branches (or however many options there are for each accessory). Then each decision feeds into the next decision point, accessory two. Obviously this pattern repeats all the way up to option 15.

number of options^number of choices n^k
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