Math problem for kindergarten

Question

Ms bullock has 6 hats . Some are red some are blue hoe many different combinations of hats can she have ?

Answer 1

15

Let R=red hat, B=blue hat

6C2, 6 objects taken 2 at a time = 6!/2!4!=15 combinations

general formula is nCr = n!/r!(n-r!) ! means multiply the given integer by every integer less that itself. 6! means 6x5x4x3x2x1=720

2! = 2x1 =2

1R5B

2R4B

3R3B

4R2B

5R1B

using all 6 hats

using just 5 hats

1R4B

2R3B

3R2B

4R1B

using 4 hats

1R3B

2R2B

3R1B

using 3 hats

1R2B

2R1B

using only 2 hats

1R1B

That's 5+4+3+2+1=15 combinations.

If all of one color are not allowed; that's the answer they're looking for: 15

as an aside, if other "combinations" are allows, such as 0R6B and 6R0B:

then add 2 more to each above, to get 7+6+5+4+3=25 combinations, but it's not really a combination of both hats, if one is missing.

The general formula is nCr = n!/r!(n-r!) in this problem n=6 and r=2

6C2 = 6!/2!4! = 6x5/2=30/2 =15

If the order of the combinations matters, that's known as permutations with a different formula

permutations distinguishes between RBBBBB and BRBBBB, etc.

6P2 or P(6,2) = 6!/4! = 30 permutations. nPr =n!/(n-r)! nPr = r!(nCr)