Question 30 of 42beginner🔧 ApplyNumerical2 marks

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.

Correct Answer

The bag contains 5 black balls and 6 red balls. We are required to select 2 black balls and 3 red balls. Since the selection of black balls is independent of the selection of red balls, we calculate the combinations separately and multiply them.

The number of ways to select 2 black balls from 5 is given by the combination formula 5C2^5C_2. Similarly, the number of ways to select 3 red balls from 6 is given by 6C3^6C_3. Following the logic of independent selection events shown in the provided examples (such as selecting red and black cards), the total number of ways is the product of these two values.

Calculation: Total ways =5C2×6C3= ^5C_2 \times ^6C_3

=5!2!⋅3!×6!3!⋅3!= \frac{5!}{2! \cdot 3!} \times \frac{6!}{3! \cdot 3!} =10×20= 10 \times 20

=200= 200.

Exercise: EXERCISE 6.4 | Q: 8 | (Page 20)
For More Understanding

Explanation

The solution applies the Fundamental Principle of Counting to two independent selection events. The provided context in Example 19(iv) demonstrates that when selecting items from distinct groups (like red and black cards), one must multiply the number of combinations for each group (26C2×26C2^{26}C_2 \times ^{26}C_2). Applying this to the balls: selecting 2 black balls from 5 is 5C2^5C_2, and selecting 3 red balls from 6 is 6C3^6C_3. The final calculation multiplies these results to get 200.

Solution Steps

  1. Step 1: Identify the total number of balls available: 5 black and 6 red.

  2. Step 2: Calculate the ways to select 2 black balls from 5 using 5C2=5!2!3!=10^5C_2 = \frac{5!}{2!3!} = 10.

  3. Step 3: Calculate the ways to select 3 red balls from 6 using 6C3=6!3!3!=20^6C_3 = \frac{6!}{3!3!} = 20.

  4. Step 4: Multiply the results to find the total number of ways: 10×20=20010 \times 20 = 200.

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