Question

Out of 800 boys in a school 224 plate cricket 248 hockey and 336 played basketball of the total 64 play basketball and hockey 80 played cricket and basketball and 40 plate cricket and hockey 24 played both

Answer 1

152

Answer 2

The problem involves calculating the number of elements outside the union of three sets, given the sizes of the individual sets and their intersections. This is a classic application of the Principle of Inclusion-Exclusion.

1. Interpret the Given Information and Assumptions: Let N be the total number of boys in the school. Let C be the set of boys who played cricket. Let H be the set of boys who played hockey. Let B be the set of boys who played basketball.

Given:

• Total boys, $N = 800$
• Number of boys who played cricket, $n(C) = 224$
• Number of boys who played hockey, $n(H) = 248$
• Number of boys who played basketball, $n(B) = 336$
• Number of boys who played basketball and hockey, $n(B \cap H) = 64$
• Number of boys who played cricket and basketball, $n(C \cap B) = 80$
• Number of boys who played cricket and hockey, $n(C \cap H) = 40$
• The phrase "24 played both" is ambiguous. Comparing with the similar question, which states "24 played all the three games", and to avoid contradiction with other given pairwise intersections, we assume "24 played both" means the number of boys who played all three games: $n(C \cap H \cap B) = 24$.
• The question is incomplete as it doesn't explicitly ask what to find. Based on the similar question, we assume it asks for the number of boys who did not play any game.

2. Apply the Principle of Inclusion-Exclusion: The number of boys who played at least one game is given by the formula for the union of three sets: $n(C \cup H \cup B) = n(C) + n(H) + n(B) - [n(C \cap H) + n(C \cap B) + n(H \cap B)] + n(C \cap H \cap B)$

Substitute the given values into the formula: $n(C \cup H \cup B) = 224 + 248 + 336 - [40 + 80 + 64] + 24$ First, sum the individual sport players: $224 + 248 + 336 = 808$ Next, sum the pairwise intersections: $40 + 80 + 64 = 184$ Now, substitute these sums back into the formula: $n(C \cup H \cup B) = 808 - 184 + 24$ $n(C \cup H \cup B) = 624 + 24$ $n(C \cup H \cup B) = 648$

3. Calculate the Number of Boys Who Did Not Play Any Game: The number of boys who did not play any game is the total number of boys minus the number of boys who played at least one game: $\text{Number of boys who did not play any game} = N - n(C \cup H \cup B)$ $\text{Number of boys who did not play any game} = 800 - 648$ $\text{Number of boys who did not play any game} = 152$

The number of boys who did not play any game is 152.