# hw5-6 - OR 2 8 C(8, 0) C(8, 1) **This is total outcomes...

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Math 1160 – Section 5.6 Homework Answer Key 4a. C(12, 4) = 495 4b. C(8, 4) = 70 4c. C(8, 2) x C(4, 2) = 168 Multiply because both must be true (2 red balls AND 2 white balls) 4d. C(8, 3) x C(4, 1) + C(8, 4) x C(4, 0) = 294 (3 red balls AND 1white ball) OR (4 red balls AND 0 white balls) 4e. C(8, 4) x C(4, 0) + C(8, 3) x C(4, 1) + C(8, 1) x C(4, 3) + C(8, 0) x C(4, 4) = 327 (4 red AND 0 white) OR (3 red AND 1 white) OR (1 red AND 3 white) OR (0 red AND 4 white) 6a. 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 (or 2 8 ) 6b. C(8, 3) = 56 6c. C(8, 2) + C(8, 3) + C(8, 4) + C(8, 5) + C(8, 6) + C(8, 7) + C(8, 8) = 247
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Unformatted text preview: OR 2 8 C(8, 0) C(8, 1) **This is total outcomes minus those not meeting criteria 6d. C(8, 4) + C(8, 5) = 126 12. Calculate the number of combinations of getting 3 answers correct OR 4 answers correct OR 5 answers correct. C(5,3) + C(5,4) + C(5,5) = 10 + 5 + 1 = 16 ways 16. Calculate the number of combinations of choosing 2 males out of the 4 males available AND choosing 2 females out of the 5 females available. C(4, 2) x C(5, 2) = 6 x 10 = 60 ways 20. C(10, 7) = 120 ways...
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## This note was uploaded on 05/21/2008 for the course MATH 116 taught by Professor Copeland during the Spring '08 term at Western Michigan.

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