Question

0.7 A bag contains 5 black balls and 6 red balls. Determine the number of ways in which 2 black and 3
red balls can be selected​

Answer 1

Answer:

There are 5 black and 6 red balls in the bag.

There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black ball in 5C2 ways and 3 red ball can be selected out of 6 red balls in 6C3 ways.

There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black ball in 5C2 ways and 3 red ball can be selected out of 6 red balls in 6C3 ways.Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls is  5C2×6C3

There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black ball in 5C2 ways and 3 red ball can be selected out of 6 red balls in 6C3 ways.Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls is  5C2×6C3=2!3!5!×3!3!6!

There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black ball in 5C2 ways and 3 red ball can be selected out of 6 red balls in 6C3 ways.Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls is  5C2×6C3=2!3!5!×3!3!6!=25×4×3×2×16×5×4=10×20=200

There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black ball in 5C2 ways and 3 red ball can be selected out of 6 red balls in 6C3 ways.Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls is  5C2×6C3=2!3!5!×3!3!6!=25×4×3×2×16×5×4=10×20=200

Answer 2

Answer:

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