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Combinatorics Difficulty 4.7 AIME Find the answer Canada

There are $28$ balls in a bag.
Each ball is coloured $1$ of $7$ colours and has $1$ of $4$ patterns. No two balls have the same
colour and pattern combination. Exactly $3$ balls are removed from the bag, one at
a time without replacement. What is the probability that one of the last
two balls removed matches the colour of the first ball and the other
matches the pattern of the first ball?

Pick one

Solution

No matter which ball is drawn first, there will be exactly $3$ balls remaining with the same colour,
and exactly $6$ balls remaining with
the same pattern. There is no overlap between these sets of $3$ balls and $6$ balls since there is only one ball with
each possible colour and pattern combination.

There are $3\times 6=18$ possible
ways for the second ball to match the colour of the first ball and the
third ball to match the pattern of the first ball.

There are $6\times 3=18$ possible
ways for the second ball to match the pattern of the first ball and the
third ball to match the colour of the first ball.

Therefore, there are $2\times 18=36$
possible ways that the remaining two balls can be drawn so that the
condition is satisfied.

There are $28$ balls in total, so
there are $27$ possibilities for the
second ball, followed by $26$
possibilities for the third ball once the second ball has been drawn.
This gives a total of $27\times 26$
ways that the last two balls can be drawn.

Putting things together, the probability that one of the last two
balls will match the colour and the other will match the pattern is
$\$\dfrac{36}{27\times}$ 26} =
$\dfrac{2}{39}\$.$

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.