Question

Assuming the balls to be identical except for difference in colors the number of ways in which one or more balls can be selected from $10$ white, $9$ green and$7$ black balls is

Answer 1

Number of ways in which one or more can be selected

$10$ white balls, $9$ green balls, and$7$ black balls are all the same.

To determine the number of possible ways to choose at least one ball.

Since,

all white balls are mutually similar, the number of ways to choose zero or more white bails is $(10+1)$

There are nine different ways to choose zero or more

The number of ways to choose zero or more black balls is $(7+1)$

As a result, the number of ways to select zero or more balls of any color is equal to $(10+1)(9+1)(7+1)$

Also, there are a total of $880$ methods to choose a total of $0$balls.

the number of possible methods to choose at least one ball (of any color) is:

$(10+1)(9+1)(7+1)-1=879.$

The number of possible methods to choose at least one ball (of any color) is $879$