Question

A bag contains 2 white marbles, 3 black marbles and 4 red marbles. Find in how many ways, 3 marbles can be drawn, so that at least one black marble is included in each draw?

Answer 1

Answer:

56 ways

Step-by-step explanation:

1 (first has to be black) * 8 (there are 9 total marbles out of which one black has been chosen) * 7 (one more has been chosen) = 56

Answer 2

Answer:

The number of ways are 64.

Step-by-step explanation:

Given : A bag contains 2 white marbles, 3 black marbles and 4 red marbles.

To Find : In how many ways, 3 marbles can be drawn, so that at least one black marble is included in each draw?

Solution :

According to question,

From 2 white marbles, 3 black marbles and 4 red marbles,.

3 marbles are to be selected such that  at least one black ball should be there.

So, Number of choices were

All three are black  - $^3C_3$

Two are black and one is non black  - $^3C_2\times ^6C_1$

One is black and two are non black  - $^3C_1\times ^6C_2$

Total number of ways  are

$^3C_3+^3C_2\times ^6C_1+^3C_1\times ^6C_2$

$=1+(3\times6)+(3\times (\frac{6\times5}{2\times1}))$

$=1+18+45$

$=64$

Therefore, The number of ways are 64.