There are 150 students in a class. The number of students who play Cricket, Hockey and Basketball are 125, 130 and 135 respectively. - Aptitude Overflow

There are 150 students in a class. The number of students who play Cricket, Hockey, and Basketball are 125, 130, 135 respectively. If 5 students do not play any of the three games, the number of students playing all the three games must be at least

In a class out of 150 studs participated in various games foot ball ,basket ball, cricket are 120, 130,135 respecly and 5 stud are not playing any one of then then find the least no of studs which are are playing all the games
a)110 b)100 c) 96

We have 125 Cricket Players, 130 Hockey Players & 135 Basketball Players.

Thus we have total (125 + 130 + 135) = 390 game players.

In the class we have 145 Player students & 5 non player students.

Each of these 145 Player students plays at least one of these 3 games( otherwise he/she would come in non player's category),

Now out of these 145 Players students we will have some 1 game players, some 2 game players and some 3 game players.

That is each of the player student can be either a 1 game players, or a 2 game player or a 3 game player.

Also all of these 145 players students will make up 390 game players.

Now suppose if all 145 students are 1 game players, then we will have 145 x 1 = 145 game players.

suppose if all 145 students are 2 game players, then we will have 145 x 2 = 290 game players,

suppose if all 145 students are 3 game players, then we will have 145 x 3 = 435 game players.

It is clear that to make 390 game players out of 145 player students, we need some 3 game player students also, because even if all of 145 students were 2 game players then also we can have only 145 x 2 = 290 players at most, and we are still 100 game players away from given 390 players hence we need at least (390 - 290) = 100, 3 game player students out of 145 player students using pigeon - hole principle.

It is a simple but highly useful principle,it states that if n items are put into m containers, with n > m, then at least one container must contain more than one item.
For Example : If you pick five cards from a standard deck of 52 cards, then at least two will be of the same suit. Since each of the five cards can belong to one of four suits and by the pigeonhole principle, two or more must belong to the same suit.