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The rules for the University of Bombay five-a-side cricket competition specify that the members of each team must have birthdays in the same month. What is the minimum number of mathematics students needed to be enrolled in the department to guarantee that they can raise a team of students?

  1. $23$
  2. $91$
  3. $60$
  4. $49$
  5. None of the above
in Combinatory edited by
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If there are n pigeon-holes and k-pigeons then some pigeon-hole contain atleast $\lceil \frac{k}{n} \rceil$ pigeons

$n=12,k=?$

$\lceil \frac{k}{12} \rceil=5$

$k=49$

 

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Take $4$ members from every month i.e $4*12 = 48$ member and now take $1$ random member which will ensure $5$ members from same month. Hence $48+1 = 49$
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3 Answers

28 votes
28 votes
Best answer

There are ${12}$ months and we have to get $5$ people having birthdays in the same month in order to form a team. Pigeon hole principle can be applied here :

$\lceil \frac{N}{12} \rceil= 5$,

$N=49 .$

Hence answer is D

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4 Comments

just adding some more points...... it can be silly but happen many times

1- In que they are saying that , they will choose 5 students from maths department randomly (its not like we are going to choose and answering that only 5 students are necessary and arguing that yes they all can born in feb or any month ) here we are not choosing either month or students.. it will choose randomly (any month any 5).

Now let us say according to @just bhavana. 12 months if every student DOB month is different and we need 5 of same months so minimum no of students are  12*4=48 now we are in a position that each month have 4 student now we need only 1 to fulfill the requirment. so 48+1 = 49 is answer
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@just_bhavana That's exactly what pigeonhole principle states.

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@just_bhavanaAwesome exp

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6 votes
6 votes
Jan Feb March April May June July August September October November December
${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$   ${\color{DarkRed} S}$   ${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$ ${\color{DarkRed} S}$
${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$   ${\color{DarkOrange} S}$ ${\color{DarkOrange} S}$
${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$   ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$ ${\color{Green} S}$
${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$ ${\color{Magenta} S}$   ${\color{Magenta} S}$ ${\color{Magenta} S}$   ${\color{Magenta} S}$

Procedure:-

Select $12$ students and then put them in slots of $1^{st}$ row each like $1^{st}$ in jan, $2^{nd}$ in feb ,$3^{rd}$ in march and so on.

Repeat the above step $4$ times.

so  Total students selected till now $= 12*4 =48$ and in each month we have $4$ students.

now if we select $1$ more student then his birthday can be in any of the given $12$ months.

say his birthday is in December then we can make a team of $5$ students such that

the members of the team must have birthdays in the same month.

$\therefore$ Total students required $= 48+1 =49.$

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0 votes
0 votes

This way we can do by PH principle

Answer:

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