There are 21 marbles in a bag. The number of each colour of marble is shown in the following table:

| Colour | Number |
| :---: | :---: |
| magenta | 1 |
| puce | 2 |
| cyan | 3 |
| ecru | 4 |
| aquamarine | 5 |
| lavender | 6 |

For example, the bag contains 4 ecru marbles. Three marbles are randomly drawn from the bag without replacement. The probability that all three of these marbles are the same colour can be written as $\frac{1}{k}$. What is the value of $k$ ?
There are $\binom{21}{3}=\frac{21(20)(19)}{3!}=7(10)(19)$ ways of choosing 3 of the 21 marbles from the bag. We count the number of ways in which 3 marbles of the same colour can be chosen.

There are 0 ways of choosing 3 magenta or 3 puce marbles, since there are fewer than 3 marbles of each of these colours.

There is 1 way of choosing 3 cyan marbles since there are exactly 3 cyan marbles in total.

Since there are 4 ecru marbles, there are $\binom{4}{3}=4$ ways of choosing 3 ecru marbles.

Since there are 5 aquamarine marbles, there are $\binom{5}{3}=\frac{5(4)(3)}{3(2)(1)}=10$ ways of choosing 3 aquamarine marbles.

Since there are 6 lavender marbles, there are $\binom{6}{3}=\frac{6(5)(4)}{3(2)(1)}=20$ ways of choosing 3 lavender marbles.

Thus, there are $1+4+10+20=35$ ways of choosing 3 marbles of the same colour.

Since there are $7(10)(19)$ ways of choosing 3 marbles, then the probability that all three are the same colour is $\frac{35}{7(10)(19)}=\frac{7(5)}{7(10)(19)}=\frac{1}{2(19)}=\frac{1}{38}$.

For this probability to equal $\frac{1}{k}$, we must have $k=38$.

ANSWER: 38