An urn contains 4 white balls and 7 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 11 balls: (a) If 2 balls are white and 1 is red? (b) If all 3 balls are white? (c) If all 3 balls are red?

(a) 2 white & 1 red ball = 42 ways

(b) All 3 white balls =  4 ways

(c) All 3 red balls = 35 ways

Step-by-step explanation:

An urn contains 4 white balls and 7 red balls. Three balls are selected.

In how many ways can the 3 balls be drawn from a total of 11 balls:

(a) If 2 balls are white and 1 is red?

The 2 white balls can be selected in ⁴C₂ ways

The 1 red ball can be selected in ⁷C₁ ways

2 white & 1 red ball = ⁴C₂ × ⁷C₁

2 white & 1 red ball = 6 × 7

2 white & 1 red ball = 42 ways

(b) If all 3 balls are white?

The 3 white balls can be selected in ⁴C₃ ways

The 0 red ball can be selected in ⁷C₀ ways

All 3 white balls = ⁴C₃ × ⁷C₀

All 3 white balls =  4 × 1

All 3 white balls =  4 ways

(c) If all 3 balls are red?

The 3 red balls can be selected in ⁷C₃ ways

The 0 white ball can be selected in ⁴C₀ ways

All 3 red balls = ⁷C₃ × ⁴C₀

All 3 red balls = 35 × 1

All 3 red balls = 35 ways

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

4 candies

step-by-step explanation:

7 x 28 = 196

200 - 196 = 4 candies remaining.