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Can you solve this?

C

Cha0s

Banned
You have 4 different blue crayons and 3 differnt red. Your pencil case can only fit 3 maximum.
You want to include at least 1 blue crayon, how many choices can you make?
**each crayon is distinguishable

Please show your logic

 
I'd grab my zippo & melt all 4 crayons, I won't have to compromise then! 😀
 
Since order doesn't matter:

P(atleast 1 blue crayon) = 1 - P(no blue crayons)

There's only 1 combination that allows no blue crayons out of 7 Cr 3 possibilities.

7!/(3!4!)

So you have 35 total combinations, so there are 34 combinations for atleast one blue crayon.

I think I did that right...

or maybe im thinking about this too complicated... 🙁
 
Originally posted by: neutralizer
Since order doesn't matter:

P(atleast 1 blue crayon) = 1 - P(no blue crayons)

There's only 1 combination that allows no blue crayons out of 7 Cr 3 possibilities.

7!/(3!4!)

So you have 35 total combinations, so there are 34 combinations for atleast one blue crayon.

I think I did that right...

or maybe im thinking about this too complicated... 🙁
Click to expand...


isn't 1 blue and 2 reds the same as 2 reds and 1 blue? But I understand how your thinking.
 
Originally posted by: Cha0s
noone got the right answer so far
Click to expand...


if this is discrete math, i'd help you, but i cannot find my book b/c i think i may have burned it by accident.


edit: ALgebra, dam, i looked way too much into this problem.
 
Originally posted by: Heisenberg
Originally posted by: Gibson486
discrete math?
Click to expand...
More like do-my-algebra-homework-for-me math. 😛
Click to expand...

He speaks truth.Its not Discrete Homework, because its to far along in the year to be working on something trivial.
 
This is discrete mathematics. And a combination problem I believe.

My guess is 180. You have 1 blue crayon that must be in one of the three slots so you have 3 nCr 1 = 3 choices to stick the blue crayon in. However since they are distinguishable from each other than you have to mulitply that by 4 for each different blue crayon, so we have 12 so far. Now removing that choice from the rest we are left with 3 blue and 3 red crayons to fit into 2 slots. So we have 6 nCr 2 = 15 choices to make. So if we multiply that by the 12 other choices we have so far we end up with 180 different combinations.

Don't know if this is right but it was worth a shot.
 
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