Cfrgtkky

So suppose you have 7 red balls, 7 blue balls, and 7 white balls, what is the probability of choosing 4 balls of one colour and the last ball of another colour without replacement.
Apparently the answer is $$frac{binom{3}{1}timesbinom{7}{4}timesbinom{2}{1}timesbinom{7}{1}}{binom{21}{5}}$$
However, I thought you could get an equivalent answer by computing $$frac{6}{20}cdotfrac{5}{19}cdotfrac{4}{18}cdotfrac{14}{17}$$ But you get twice the probable chance from the choose method. Could somebody explain why the second way is flawed?

I would question the validity of the answer involving binomial coefficients...

Surely, the total number of ways of choosing $5$ balls from $21$ is $binom{21}{5}$, so as long as the rest of our counting argument is fine, then after diving by this number we will recover the probability we desire (since everything is uniformly random).

Whoever wrote this answer seems to say that we should "choose a color", then "draw four balls of that color", then "choose a different color", and finally "draw one ball of that color". But this is a touch shady, because the exact same expression with the numerator reordered would have you choose a color of which to draw a single ball, and then choose a different color of which to draw $4$ balls!

Since your stated problem seems to require that you draw the odd ball out last, the answer involving binomial coefficients is wrong because it is blind to order; specifically, it counts twice the number of scenarios that it should as valid.

Your answer is correct.
The other answer is actually $5$ times as large as it should be.

What the other answer gave is the probability that among $5$ balls chosen from $21,$
$4$ balls will be chosen from one of the $7$ balls of one color
(which gives the factor $binom74$),
that color can be any of the original three (whence the factor $binom31$),
and $1$ ball will be chosen from one of the $7$ balls (whence the factor $binom71$)
of one of the remaining $2$ colors (whence the factor $binom21$).

Notice, however, that in the preceding paragraph I said nothing about the singleton color ball being chosen last. In fact the answer you were given counts the number of ways to draw four balls of one color and one of another where the singleton ball could occur anywhere in the sequence of balls drawn.
That's because the whole idea of counting events from $binom{21}{5}$ possibilities
is that each of these events describes $5$ balls chosen from the urn without regard to the order in which they were chosen.
So it is counting all the events where the singleton ball was the first, second, third, or fourth ball chosen, not just the last.

In fact, it should be a huge red flag (indicating a likely error)
to see $binom{21}{5}$ in the denominator of the alleged probability of drawing five ballse from an urn of $21$ balls such that the last ball is a different color from the others.