TITLE: A problem on distributing 29 disks on $7\times 7$ grid
QUESTION [6 upvotes]: I got this problem:
Given a $7\times 7$ grid, if we distribute $29$ disks on the grid such that each square cannot hold more than $1$ disk, what is the probability that there will be at least one row full of disks on the grid?
My first try:
$P(\{\text{there is at least one row full of disks}\}= \frac{7\times{42\choose 22}}{49\choose 29}$
Since we have $7$ ways to choose the row that we will fill by disks, and then we have remaining $22$ disks which we will distribute over the remaining $42$ squares. But this is obviously wrong since we count some combinations multiple times.
My second try:
$P(\{ \text{there is at least one row full of disks}\}= P(\{\text{there is exactly 1 row full of disks}\}\cup\{\text{there is exactly 2 rows full of disks}\}\cup\{\text{there is exactly 3 rows full of disks}\}\cup\{\text{there is exactly 4 rows full of disks}\})=P(\{\text{there is exactly 1 row full of disks}\}+P(\{\text{there is exactly 2 rows full of disks}\}+P(\{\text{there is exactly 3 rows full of disks}\}+P(\{\text{there is exactly 4 rows full of disks}\}$
But this probably makes things harder and does not simplifies things.
My third try:
$P(\{\text{there is at least one row full of disks}\}= 1-P(\{\text{there are no rows full of disks}\})$
But I got stuck, I tried to count the number of combinations in which each row got an empty square but here too I counted some combinations multiple times.
Any hint/help will be appreciated.

REPLY [3 votes]: There are ${49\choose20}={49\choose29}$ equiprobable ways to choose the  cells obtaining no disk, called empty cells in the following. Each such choice constitutes an arrangement.
We have to count the number $N$ of arrangements where at least one row contains no empty cell. For this count we need the the inclusion-exclusion-principle. It gives
$$N={7\choose1}{42\choose 20}-{7\choose2}{35\choose 20}+{7\choose3}{28\choose 20}-{7\choose4}{21\choose 20}=3528443228520\ .\tag{1}$$
(As an example consider the number of arrangements with exactly three rows containing no empty cell. These are counted $3$ times in the first term of $(1)$, $-3$ times in the second term, and $1$ time in the third term.)
The probability $p$ in question is therefore given by
$$p={N\over{49\choose20}}={3000376895\over24045516451}\doteq0.124779\ .$$