Question

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is
(a) 879
(b) 880
(c) 629
(d) 630

Answer 1

Hint: To solve this question we use a concept of selection of some elements from the given set of elements. Total number of ways in which a selection can be done by taking some or all out of \[(p+q+r)\] items where p are alike, q are alike, r alike is given by \[[(1+p)(1+q)(1+r)-1]\].

Complete step-by-step answer:
Similarly if we are given to determine or select more than (p+q+r) items from the set or (p+q+r…..) elements from the set then the number of ways to do so is \[[(1+p)(1+q)(1+r)........-1]\]

We are given in the question as 10 White, 9 Green and 7 Black balls.
To use the formula as stated let us assume p=10 White balls, q=9 Green balls and r=7 Black balls.
Now here in the question we are only given to determine the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls, that is, the number of ways in which one or more balls can be selected from p, q are r items.
We have, total number of ways in which a selection can be done by taking some or all of \[(p+q+r)\]items where p are alike , q are alike , r alike is given by \[[(1+p)(1+q)(1+r)-1]\].
Substituting the values of p=10, q=9 and r=7 we get,
\[[(1+p)(1+q)(1+r)-1]=[(1+10)(1+9)(1+7)-1]\]
\[\Rightarrow [(1+p)(1+q)(1+r)-1]=[(11)(10)(8)-1]\]
\[\Rightarrow [(1+p)(1+q)(1+r)-1]=[880-1]\]
\[\Rightarrow [(1+p)(1+q)(1+r)-1]=879\]
Hence, the total number of ways in which a selection can be done by taking some or all of \[(p+q+r)\]items where p are alike, q are alike, r alike is given by 879.

Therefore, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is 879.

Note: The possibility of mistake is that we can assume all balls of every colour to be distinct or different from each other, it will give a wrong implemented formula, which is not the case because given is that all the balls of same colour are identical and difference in between them is only considered only with respect to colour. There is also an alternative method to do this question which is given by selecting r out of n items as \[{}_{r}^{n}C=\dfrac{n!}{(r!)(n-r)!}\]