Of the members of three athletic teams in a certain school 21 are on the basket ball team, 26 on the hockey team, and 29 on the football team, 14 play hockey and basketball. 15 play hockey and football, and 12 play football and basketball, 8 are on all the three teams. How many members are there altogether?

Options

A. 38
B. 47
C. 51
D. 43

Answer

To solve the problem, we will use the principle of inclusion-exclusion to find the total number of distinct members in the three athletic teams. Let's denote:

- \( n(B) \): Number of members in the basketball team = 21
- \( n(H) \): Number of members in the hockey team = 26
- \( n(F) \): Number of members in the football team = 29
- \( n(H \cap B) \): Number of members in both hockey and basketball = 14
- \( n(H \cap F) \): Number of members in both hockey and football = 15
- \( n(F \cap B) \): Number of members in both football and basketball = 12
- \( n(B \cap H \cap F) \): Number of members in all three teams = 8

We want to find the total number of distinct members in the union of the three teams, denoted as \( n(B \cup H \cup F) \).

Using the principle of inclusion-exclusion, we can express this as:

\[ n(B \cup H \cup F) = n(B) + n(H) + n(F) - n(H \cap B) - n(H \cap F) - n(F \cap B) + n(B \cap H \cap F) \]

Now, substituting the values we have:

\[ n(B \cup H \cup F) = 21 + 26 + 29 - 14 - 15 - 12 + 8 \]

Calculating step by step:

1. Calculate the sum of members in each team:
\[ 21 + 26 + 29 = 76 \]

2. Calculate the sum of members in the intersections of two teams:
\[ 14 + 15 + 12 = 41 \]

3. Now substitute these sums into the equation:
\[ n(B \cup H \cup F) = 76 - 41 + 8 \]

4. Calculate:
\[ 76 - 41 = 35 \]
\[ 35 + 8 = 43 \]

Thus, the total number of distinct members in all three teams is \( \boxed{43} \).