This is a problem based on pigeon hole principle.
A tennis player has three weeks to prepare for a tennis tournament. She decides to play at least one set every day but not more than 36 in all. Show that there is a period of consecutive days during which she will play exactly 21 sets.
Can some one please help me to solve this problem.
Best Answer
Let $r_i$ be the number of sets played up to and including day $i$. So
$0 = r_0 < r_1 < r_2 < ... < r_{21} \leq 36$
So there are 37 pigeonholes - possible values - for the 22 pigeons - the $r_i$.
However, we are looking for pairs $i, j$ with $r_j = r_i + 21$ - since such a pair would tell us that 21 sets were played on days $i+1$ through to $j$. So if we assume there are no such pairs, we can "combine" the pigeonholes $i$ and $i + 21$. So we have the following pigeonholes:
$\{0 \text{ or } 21\}, \{1 \text{ or } 22\}, ..., \{15 \text{ or } 36\}, \{16\}, ... \{20\}$.
And we have 22 pigeons, 21 pigeonholes, giving a contradiction.