Question

A box contains 5 white balls and 7 black balls. In how many ways can 3 balls be drawn from the box such that at least one black ball is included?}

• A. 210
• B. \textbf{175}
• C. 133
• D. 70

Hint:

To count “at least one” cases, subtract the unwanted (all white) from the total combinations. This is often faster than summing each possible case.

Answer 1

The Correct Option is B

Total number of balls = \(5 + 7 = 12\) Total number of ways to choose any 3 balls from 12: \[ \binom{12}{3} = 220 \] We need the number of ways to choose 3 balls such that **at least one is black**. So, we subtract the number of ways of choosing **only white balls** from the total: Number of ways to choose 3 white balls out of 5: \[ \binom{5}{3} = 10 \] Thus, number of ways to choose at least one black ball: \[ \binom{12}{3} - \binom{5}{3} = 220 - 10 = 210 \] Oops! But this contradicts the answer (175). Let’s re-check. Wait! There seems to be a mistake here — the answer is **175**, not 210. That suggests the question might be **selecting 3 balls where all combinations with at least one black ball are valid**, but possibly **without replacement** and **no order**. Let’s list the favorable cases with at least one black: - Case 1: 1 white, 2 black \(\binom{5}{1} \cdot \binom{7}{2} = 5 \cdot 21 = 105\) - Case 2: 2 white, 1 black \(\binom{5}{2} \cdot \binom{7}{1} = 10 \cdot 7 = 70\) - Case 3: 3 black \(\binom{7}{3} = 35\) Total = \(105 + 70 + 35 = \boxed{210}\) Wait again! This gives 210. That matches our original count, meaning the **correct answer is actually (1) 210**, not 175. So, the scanned answer (175) is likely **incorrect**. Let’s double check: - All 3 white balls: \(\binom{5}{3} = 10\) - All 3 combinations: \(\binom{12}{3} = 220\) So, combinations with at least one black: \[ 220 - 10 = \boxed{210} \] Hence, correct answer: **(1) 210** Let’s update the answer accordingly. % Correct Answer (Updated) % Correct Answer Correct Answer:} (1) 210