Six fair coins are tossed at the same time; in how many different outcomes do at most three of the coins show heads?

Introduction / Context:
This question checks your knowledge of binomial coefficients in the context of coin tossing. When multiple fair coins are tossed simultaneously, each outcome can be described by the number of heads, and counting outcomes with restrictions such as at most three heads is a standard combinatorics task in aptitude exams.

Given Data / Assumptions:

• Six identical fair coins are tossed at the same time.
• Each coin has two outcomes: Head (H) or Tail (T).
• We must count the number of outcomes in which the total number of heads is at most three.
• At most three heads means 0, 1, 2 or 3 heads are allowed.

Concept / Approach:
The number of outcomes with exactly k heads when n fair coins are tossed is given by the binomial coefficient nCk, because we only need to choose which k of the n coins show heads. To find the outcomes with at most three heads, we sum nCk for k = 0, 1, 2 and 3 when n = 6.

Step-by-Step Solution:
Total coins n = 6.We want sum of 6C0 + 6C1 + 6C2 + 6C3.Compute: 6C0 = 1 (all tails).6C1 = 6, 6C2 = 15, 6C3 = 20.Add them: 1 + 6 + 15 + 20 = 42.Therefore, there are 42 such outcomes.

Verification / Alternative check:
Total outcomes for six coins = 2^6 = 64.Outcomes with more than three heads are those with 4, 5 or 6 heads.Compute 6C4 = 15, 6C5 = 6, 6C6 = 1.Sum = 15 + 6 + 1 = 22.Now 64 - 22 = 42, which matches our previous result.

Why Other Options Are Wrong:
15 is just the value of 6C2 and does not include other allowed cases.16 could result from wrongly assuming half the outcomes qualify.40 is near the correct value but misses two outcomes due to miscounting.

Common Pitfalls:
Forgetting to include the case with 0 heads.Using permutations instead of combinations when counting head positions.Misreading at most three heads as exactly three heads.

Final Answer:
The number of outcomes with at most three heads is 42.