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Can you solve this?

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Originally posted by: 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
Click to expand...

Piece of cake.
Simple counting principle problem.
if the color didn't matter, there would be 7*6*5 / (3*2*1) different ways to choose 3 crayons.
(combination of 3 out of 7)
That's 35 different combinations of 3 crayons.

There's only 1 combination where there isn't at least 1 blue crayon. (choose all 3 red ones) So, 35 - 1 = 34.
 
(oh, and neutralizer had it right... I didn't read the thread before posting)
 
now if you put them into actual boxes and number each you get the same answer as above, however no one has accounted for the boxes that aren't full

which would include 4 for each of the boxes that have only 1 blue crayon in it
and another 5 which only have 2 blue crayons in it.

all combinations would be if you make b1, b2, b3, b4 each of the blue crayons
and r1, r2, r3 each of the red crayons:

non full boxes:
b1
b2
b3
b4

b1, b2
b1, b3
b1, b4
b2, b3
b2, b4

= 9 boxes total non full with at least 1 blue crayon

full boxes:

(ALL BLUE)
b1, b2, b3
b1, b2, b4
b2, b3, b4
b1, b3, b4
= 4 boxes total full with 3 blue crayons

(1 BLUE, 2 REDS)
b1, r1, r2
b2, r1, r2
b3, r1, r2
b4, r1, r2

b1, r1, r3
b2, r1, r3
b3, r1, r3
b4, r1, r3

b1, r2, r3
b2, r2, r3
b3, r2, r3
b4, r2, r3
= 12 boxes total full with 1 blue and 2 red crayons

(2 BLUE, 1 RED):
b1, b2, r1
b1, b2, r2
b1, b2, r3

b1, b3, r1
b1, b3, r2
b1, b3, r3

b1, b4, r1
b1, b4, r2
b1, b4, r3

b2, b3, r1
b2, b3, r2
b2, b3, r3

b2, b4, r1
b2, b4, r2
b2, b4, r3

b3, b4, r1
b3, b4, r2
b3, b4, r3
= 18 boxes total full with 2 blue and 1 red crayons

OVERALL TOTAL of 34(full boxes) = (7C3) + 9 (non full boxes) == 43 BOXES with at least 1 BLUE CRAYON in it.
 
Originally posted by: ocard
now if you put them into actual boxes and number each you get the same answer as above, however no one has accounted for the boxes that aren't full

which would include 4 for each of the boxes that have only 1 blue crayon in it
and another 5 which only have 2 blue crayons in it.

all combinations would be if you make b1, b2, b3, b4 each of the blue crayons
and r1, r2, r3 each of the red crayons:

non full boxes:
b1
b2
b3
b4

b1, b2
b1, b3
b1, b4
b2, b3
b2, b4

= 9 boxes total non full with at least 1 blue crayon

full boxes:

(ALL BLUE)
b1, b2, b3
b1, b2, b4
b2, b3, b4
b1, b3, b4
= 4 boxes total full with 3 blue crayons

(1 BLUE, 2 REDS)
b1, r1, r2
b2, r1, r2
b3, r1, r2
b4, r1, r2

b1, r1, r3
b2, r1, r3
b3, r1, r3
b4, r1, r3

b1, r2, r3
b2, r2, r3
b3, r2, r3
b4, r2, r3
= 12 boxes total full with 1 blue and 2 red crayons

(2 BLUE, 1 RED):
b1, b2, r1
b1, b2, r2
b1, b2, r3

b1, b3, r1
b1, b3, r2
b1, b3, r3

b1, b4, r1
b1, b4, r2
b1, b4, r3

b2, b3, r1
b2, b3, r2
b2, b3, r3

b2, b4, r1
b2, b4, r2
b2, b4, r3

b3, b4, r1
b3, b4, r2
b3, b4, r3
= 18 boxes total full with 2 blue and 1 red crayons

OVERALL TOTAL of 34(full boxes) = (7C3) + 9 (non full boxes) == 43 BOXES with at least 1 BLUE CRAYON in it.
Click to expand...

You created an account just to post in this thread?

wwybywb?
 
not precisely.. I like to solve problems. It wasn't stated explicitly that the boxes had to be full. Therefore no one actually had the correct answer. 🙂
 
Originally posted by: ocard
not precisely.. I like to solve problems. It wasn't stated explicitly that the boxes had to be full. Therefore no one actually had the correct answer. 🙂
Click to expand...

So you randomly happened by OT looking for a math thread?
 
yes, well not completely random. I've never been to your website before until I talked to a friend of mine who also posts messages here. Then I just went to that one for different topics. Hope I didn't offend you.
 
but your explanation does not account for those boxes which aren't full
Mister I take probability and statistics
Click to expand...
, because I'm sure you aren't the only person in the world that can take that course..
 
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