Question

Find the number of ways in which one or more balls can be selected from a lot of 10 white 9 green and 7 black balls assuming balls to be identical except for the difference of colour.

a) 880

b) 629

c) 630

d) 879

Answer 1

Hint: The number of ways in which r items can be selected out of a lot of m identical things of one type, n identical things of another type and p identical things of the third type is equal to the product of the number of ways to select each particular type of thing. Also consider the case when nothing is selected from a lot and this case must be subtracted from all possible ways to select a thing. 

Complete step-by-step answer:

Consider the case of white balls, either 0 or 1,2,3….10 white balls can be selected.

So, Write the number of ways to select white ball that is 10+1 ways.

Now, consider the case of green balls, either 0 or 1,2,3….9 green balls can be selected.

So, Write the number of ways to select the green ball that is 9+1 ways.

Consider the case of black balls, either 0 or 1,2,3….7 black balls can be selected.

So, Write the number of ways to select black ball that is 7+1 ways.

Here note that there will be one case when no ball will be selected. So, the case when no ball is selected is 1, which is to be subtracted from the total number of ways to select atleast one of the balls.

So, the total number of ways to select 1 or more balls is $\left( 1+10 \right)\left( 1+9 \right)\left( 1+7 \right)-1$

= $\left( 11 \right)\left( 10 \right)\left( 8 \right)-1$

= $880-1 = 879$

Hence, option (d) is correct.

Note: Whenever such cases appear, do not forget to subtract the case when no item has been selected. Then consider the cases for each item to be selected from n different types of items. No item selection must be subtracted from the all possible number of ways where at least one item is to be selected.