Pigeon hole at least

Question

In a team there are players between $18$ until $25$ years old. How many at least it must be to take part in one excursion, as that it will be sure that will be at least 10 persons same age and same gender?

What I did like this I try solve it. I believe I am wrong; if you can help me... I take $x$ persons out of $10$ so it will give me $9$. And $9$ is the worst case so $$\left\lceil \frac x{10} \right\rceil = \lceil9\rceil \implies \lceil x \rceil = \lceil 9\cdot10 \rceil = \lceil 90 \rceil.$$ I feel I am wrong. Can you help me?

Answer 1

Your holes are elements of form (age, gender). There are (25 - 18 + 1) * 2 = 16 possible configurations. Now we use strong version of pigeon hole principle:

Let q1, q2, ..., qn be positive integers. If

$ q_{1}+q_{2}+\cdots +q_{n}- n+1$ objects are distributed into n boxes, then either the first box contains at least $q_1$ objects, or the second box contains at least $q_2$ objects, ..., or the nth box contains at least $q_n$ objects

by this theorem you can check that distributing 145 object into 16 holes can give you at lest on hole with 10 objects. If you distrubute less then 145 there exists configuration where no hole has 10 object ie. for 144 every hole could have 9 objects.