Can you solve this?

[Cha0s]

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

 

[Heisenberg]

Are the crayons distinguishable other than by their color?

 

[DaTT]

3

 

[Zim Hosein]

I'd grab my zippo & melt all 4 crayons, I won't have to compromise then!

 

[DaTT]

3 blue, 2blue 1 red, 1 blue 2 red

 

[eigen]

Originally posted by: DaTT
3 blue, 2blue 1 red, 1 blue 2 red

Yeah but is Order important?
Does he want the number of lists or the number of sets.

 

[chuckywang]

1 blue, 2 reds
2 blues, 1 red
3 blues

Three cases.

 

[Cha0s]

Originally posted by: Heisenberg
Are the crayons distinguishable other than by their color?

yes

 

[Zanix]

Break them in halves, then you can have 42 different combinations.

 

[upsciLLion]

Originally posted by: DaTT
3

Yup. You can take one blue and two red, two blue and one red, or three blue and no red.

 

[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...

 

[DaTT]

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...

isn't 1 blue and 2 reds the same as 2 reds and 1 blue? But I understand how your thinking.

 

[Gibson486]

discrete math?

 

[Cha0s]

wow, quick thinking

 

[Heisenberg]

Originally posted by: Gibson486
discrete math?

More like do-my-algebra-homework-for-me math.

 

[Gibson486]

Originally posted by: Cha0s
noone got the right answer so far

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.

 

[SurgicalShark]

34 possibilities if each one is distinguishable

 

[eigen]

Originally posted by: Heisenberg

Originally posted by: Gibson486
discrete math?

More like do-my-algebra-homework-for-me math.

He speaks truth.Its not Discrete Homework, because its to far along in the year to be working on something trivial.

 

[SurgicalShark]

4c3 + (4c2 X 3c1) + (4c1 X 3c2) = 34

 

[neutralizer]

Originally posted by: SurgicalShark
34 possibilities if each one is distinguishable

hey i said too

logic included too

atleast someone agrees with me.

 

[HonkeyDonk]

there are a lot of combos.

 

[neutralizer]

Originally posted by: HonkeyDonk
there are a lot of combos.

No, only 7 Cr 3 combos.

 

[SurgicalShark]

Originally posted by: neutralizer

Originally posted by: SurgicalShark
34 possibilities if each one is distinguishable

hey i said too

logic included too

atleast someone agrees with me.

sorry if you answered before I did, I did not read any responses.

 

[ohnnyj]

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.

 

[HonkeyDonk]

Originally posted by: neutralizer

Originally posted by: HonkeyDonk
there are a lot of combos.

No, only 7 Cr 3 combos.

which is a lot! ...relatively speaking heh.