IBM
Company
Numerical Ability
Permutation and Combination
Last month I sent off for one of those kits which you can use to make your own Christmas Crackers. The kit contained:
Three colours of hat: Red, Yellow and Blue
Four types of novelty: toy car, spinning top, magnifying glass and miniature hair brush
Four different types of joke slip
All the other parts were the same type. The kit contained enough bits for 50 crackers. Can I make each cracker different from all the others?
Read Solution (Total 5)
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- yes 3*4*4*remaining give different solutions
- 10 years agoHelpfull: Yes(6) No(0)
- 3 x 4 x 4 = 48. Sadly at least two crackers must be exact duplicates of ones already made.
BUT if you could handle the sad look on people's faces, you could consider a missing item as a possible combination. This gives 4 x 5 x 5 = 100 different crackers. 40 crackers would be missing one item, 11 will be missing two, and one cracker with nothing in at all! [COPIED] - 10 years agoHelpfull: Yes(4) No(0)
- is the answer YES or NO..???
- 10 years agoHelpfull: Yes(2) No(0)
- ans is 5 because there are 5 rows with atleast 3 chairs
- 9 years agoHelpfull: Yes(0) No(0)
- no.......
3 x 4 x 4 = 48. Sadly at least two crackers must be exact duplicates of ones already made.
BUT if you could handle the sad look on people's faces, you could consider a missing item as a possible combination. This gives 4 x 5 x 5 = 100 different crackers. 40 crackers would be missing one item, 11 will be missing two, and one cracker with nothing in at all! [COPIED]
Read more at http://www.m4maths.com/placement-puzzles.php?SOURCE=IBM&TOPIC=Numerical%20Ability&SUB_TOPIC=Permutation%20and%20Combination#vMtDubwLVbxwLMBG.99 - 9 years agoHelpfull: Yes(0) No(0)
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