Question

An urn contains $5$ red marbles, $4$ black marbles and $3$ white marbles. Then the number of ways in which $4$ marbles can be drawn so that at most three of them are red.

Answer 1

Step:1 Calculate the different cases of selecting most of the red balls.

Total red marbles in the urn are $5$ and total other marbles are: $4+3=7$.

Now, the cases to pick $4$ marbles are:

$3$ red marbles and $1$other marble.

$2$ red marbles and $2$other marble.

$1$ red marbles and $3$other marble.

$0$ red marbles and $4$other marble.

Step:2 Calculate the total number of cases.

Therefore, total cases

$$\begin{gathered} =C_3^5\times C_1^7+C_2^5\times C_2^7+C_1^5\times C_3^7+C_0^5\times C_4^7 \\ =\frac{5!}{3!\times 2!}\times 7+\frac{5!}{3!\times 2!}\times \frac{7!}{5!\times 2!}+5\times \frac{7!}{4!\times 3!}+1\times \frac{7!}{3!\times 4!} \\ =\frac{5\times 4\times 3!}{3!\times 2}\times 7+\frac{7\times 6\times 5\times 4\times 3!}{3!\times 2\times 2}+5\times \frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}+\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!} \\ =70+210+175+35 \\ =490 \end{gathered}$$

Therefore, Option(C) is the correct answer.