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Question

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

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Solution

It is given that a bag contains 5 black and 6 red balls.

Since 2 black balls are to be chosen then the number of ways of selecting them are the combination of 5 black balls taken 2 at a time.

The formula for the combination is defined as,

C n r = n! ( nr )!r!

Substitute 5 for n and 2 for r in the above formula.

C 5 2 = 5! ( 52 )!2! = 5! 3!2!

Cancel the common factors by factorizing the bigger term to the factorial.

The formula to calculate the factors of a factorial in terms of factorial itself is,

n!=n( n1 )! n!=n( n1 )( n2 )![ n2 ]

The combination is written as,

C 5 2 = 5×4×3! 3!2×1 = 5×4 2 =10

The number of ways that the black balls are selected is 10.

Since 3 red balls are to be chosen then the number of ways of selecting them are the combination of 6 black balls taken 3 at a time.

The formula for the combination is defined as,

C n r = n! ( nr )!r!

Substitute 6 for n and 3 for r in the above formula.

C 6 3 = 6! ( 63 )!3! = 6! 3!3!

Cancel the common factors by factorizing the bigger term to the factorial.

The formula to calculate the factors of a factorial in terms of factorial itself is,

n!=n( n1 )! n!=n( n1 )( n2 )![ n2 ]

The combination is written as,

C 6 3 = 6×5×4×3! 3!3×2×1 = 6×5×4 3×2×1 =20

The number of ways that the red balls are selected is 20.

By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, , the total number of ways is m×n.

The number of ways that 2 black and 3 red balls are selected is,

10×20=200

Thus, required number of ways is 200.


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