Question

38. From 60 players, playing cricket, football and hockey, each player has to play at least one

game. 45 played cricket, 30 played football and 40 played hockey. Atmost 27 played

cricket and football, atmost 32 played football and hockey and atmost 25 played cricket

and hockey. Find maximum number of players that could have played all three games.​

Answer 1

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GIVEN

• From 60 players, playing cricket, football and hockey, each player has to play at least one game.

• From 60 players, playing cricket, football and hockey, each player has to play at least one game. 45 played cricket, 30 played football and 40 played hockey.

• At most 27 played cricket and football, at most 32 played football and hockey and at most 25 played cricket and hockey.

TO DETERMINE

The maximum number of players that could have played all three games.

CALCULATION

Let C = Cricket , F = Football and H = Hockey

Then by the given condition

$\sf{n( C ) = 45 \: , n( F ) = 30 \: , n( H )= 40 \: }$

$\sf{n( C \cap \: F) \leqslant \: 27 \: , \: n( F \cap \: H ) \leqslant \: 32 \: , \: n( C \cap \: H ) \leqslant \: 25 }$

Also

$\sf{n( C \cup \: F \cup \: H ) = 60}$

We know from set theory that

$\sf{n( C \cup \: F \cup \: H ) = n( C) + \: n(F) + n( H ) \: \: }$

$\sf{ - n( C \cap \: F) - n( F \cap \: H ) - \: n( C \cap \: H ) + n( C \cap \: F \cap \: H ) }$

$\implies \sf{n( C \cap \: F \cap \: H ) = n( C \cup \: F \cup \: H ) - n( C) - \: n(F) - n( H ) \: \: }$

$\sf{ + n( C \cap \: F) + n( F \cap \: H ) + \: n( C \cap \: H ) }$

$\implies \sf{n( C \cap \: F \cap \: H ) } \leqslant 60 - 45 - 30 - 40 + 27 + 32 + 25$

$\implies \sf{n( C \cap \: F \cap \: H ) } \leqslant 29$

Hence the maximum number of players that could have played all three games are 29

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LEARN MORE FROM BRAINLY

If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B′) = 350, find n(B) and n(U).

https://brainly.in/question/4193770

Answer 2

Step-by-step explanation:

\displaystyle\huge\red{\underline{\underline{Solution}}}

Solution

GIVEN

From 60 players, playing cricket, football and hockey, each player has to play at least one game.

From 60 players, playing cricket, football and hockey, each player has to play at least one game. 45 played cricket, 30 played football and 40 played hockey.

At most 27 played cricket and football, at most 32 played football and hockey and at most 25 played cricket and hockey.

TO DETERMINE

The maximum number of players that could have played all three games.

CALCULATION

Let C = Cricket , F = Football and H = Hockey

Then by the given condition

\sf{n( C ) = 45 \: , n( F ) = 30 \: , n( H )= 40 \: }n(C)=45,n(F)=30,n(H)=40

\sf{n( C \cap \: F) \leqslant \: 27 \: , \: n( F \cap \: H ) \leqslant \: 32 \: , \: n( C \cap \: H ) \leqslant \: 25 }n(C∩F)⩽27,n(F∩H)⩽32,n(C∩H)⩽25

Also

\sf{n( C \cup \: F \cup \: H ) = 60}n(C∪F∪H)=60

We know from set theory that

\sf{n( C \cup \: F \cup \: H ) = n( C) + \: n(F) + n( H ) \: \: }n(C∪F∪H)=n(C)+n(F)+n(H)

\sf{ - n( C \cap \: F) - n( F \cap \: H ) - \: n( C \cap \: H ) + n( C \cap \: F \cap \: H ) }−n(C∩F)−n(F∩H)−n(C∩H)+n(C∩F∩H)

\implies \sf{n( C \cap \: F \cap \: H ) = n( C \cup \: F \cup \: H ) - n( C) - \: n(F) - n( H ) \: \: }⟹n(C∩F∩H)=n(C∪F∪H)−n(C)−n(F)−n(H)

\sf{ + n( C \cap \: F) + n( F \cap \: H ) + \: n( C \cap \: H ) }+n(C∩F)+n(F∩H)+n(C∩H)

\implies \sf{n( C \cap \: F \cap \: H ) } \leqslant 60 - 45 - 30 - 40 + 27 + 32 + 25⟹n(C∩F∩H)⩽60−45−30−40+27+32+25

\implies \sf{n( C \cap \: F \cap \: H ) } \leqslant 29⟹n(C∩F∩H)⩽29

Hence the maximum number of players that could have played all three games are 29

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B′) = 350, find n(B) and n(U).

https://brainly.in/question/4193770