Sunday, May 16, 2010

Problem solving with combinations?

1. in how many ways can you select 4 boys %26amp; 3 girls from 9 boys %26amp; 6 girls to sit at the head table at the school fundraiser?





2. Nickson is shopping for clothes %26amp; finds that there are 9 items in his size at the store. in how many ways can Nickson make a purchase?





3. how many different sums of money can be made with a $1 coin, a $2 coin, 4 $5 bills, and a $50 bill?





4. Linda wants to put some CDs in her car to listen to on her way to her cottage. if she has a choice of 12 soft rock, 4 classical, %26amp; 6 rap CDs, in how many ways can Linda choose some CDs to put in her car?





5. When preparing a bouquet, a florist can choose from 7 roses, 6 carnations, %26amp; 4 chrysanthemums. how many bouquets can the florist make if the bouquet must have at least one flower?





THANK YOU IN ADVANCE! :)

Problem solving with combinations?
Most of the answers have a -1 at the end to indicate you can't have 0 of any of these. If you can have zero of all items, take off the -1.





1.9c4 ways to choose the boys and 6c3 ways to select the girls. So (9c4) * (6c3)





2. He can purchase 9 or 8 or 7 or... or 1:





9c9 + 9c8 + 9c7 + ... + 9c1 =





(sum from N=0 to 9 of 9cN)





3. [1c1 + 1c0]*[1c1 + 1c0] * [4c4 + 4c3 + ... + 4c0] * [1c1 + 1c0] - 1





4. She can put [12c12 + 12c11 +...+12c0] * [4c4+4c3+...+4c0] * [6c6 + 6c5 + ... + 6c0] - 1





5. [7c7 + 7c6 + ... + 7c0] * [6c6 + 6c5 + ... + 6c0] * [4c4 + 4c3 + ... + 4c0] - [7c0*6c0*4c0].





(that last part 7c0*6c0*4c0 is just = 1 since there is only 1 way to not have any flowers.)


No comments:

Post a Comment