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+10 votes

What is the minimum number of students needed in a class to guarantee that there are at least $6$ students whose birthdays fall in the same month ?

- $6$
- $23$
- $61$
- $72$
- $91$

+14 votes

Best answer

+5

this concept something like deadlock condition....if one of the process satisfy (at least condition) then no more deadlock

0

Yes, also I don't think you have to remember any formula for this question.

It's very logical.

You have $12$ months. So if there are 5 students born in each month in the worst case then you get $\mathbf{12\times5 = 60}$ but you need a **guarantee **of at least $6$ students whose b'day fall in the same month. $\therefore$ Add $\mathbf{+1}$ more.

So answer $=\mathbf{12\times5 + 1 = 61}$

+2 votes

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**Procedure :-**

select $12$ students and then put them in $1$ slot each like $1^{st}$ Jan $, 2^{nd}$ in Feb$, 3^{rd}$ in March and so on.

repeat the above step $5$ times.

so total students selected till now $= 12\times 5 =60$ and in each month we have $5$ 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

guarantee that there are at least $6$ students whose birthdays fall in the same month

so total students required $= 60 +1 =61.$

0 votes

By pigeon hole principal we can ensure that in 61 we will have atleast 6 student having birthday on same month

n=km+1 k=5, m=12

n=km+1 k=5, m=12

0 votes

$\mathbf{\underline{Answer:C}}$

$\mathbf{\underline{Intuitive \;Answer:}}$

It's very logical.

You have $\mathbf{12}$ months. So if there are $\mathbf5$ students born in each month in the worst case then you get $\mathbf{12\times5 = 60}$ but you need a **guarantee **of at least $\mathbf 6$ students whose b'day fall in the same month.

$\therefore$ Add $\mathbf{+1}$ more.

So answer $=\mathbf{12\times5 + 1 = 61}$

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