The total number of ways of selecting one or more items from $p$ identical items of one kind, $q$ identical objects of second kind, $r$ identical objects of third kind and n different items is $(p+1)(q+1)(r+1)(2^n-1$)
Can someone tell me how this expression here?I thought number of ways of selection from identical object is always $1$.
On
Actually you have wrong answer with you. Look up for the correct one first.
Now since you say that you have p number of objects which are of same kind. And similarly q are of one type and r of one type and remaining n of different types.
Assume that you wish to choose from p type items then either you choose 0 from them or one or two or three and so on till pth one. Hence now you have p+1 ways of selecting or choosing from p family items.
Draw the same analogy for q and r too.
But since all are done at same time then its a compulsory event which implies that you need all cases altogether thus multiply all of them.
But there remains n different type of elements. And I assume that you now how 2^n has got to link with its selections.
But wait! As you have assumed that you choise 0 from p or 0 from q or 0 from r then it might also happen that during a particular selection you chose neither of all three. And we are not interested in this case. Hence we subtract 1 at the 'end'
1 For the identical sets of objects
Number of objects we can pick from the $p$ identical objects of the first kind are-
$0,1,2,3,...,p-1$ or $p$
Which are exactly $(p+1)$ distinct values
The same applies for each of the other sets of identical items.
2 For the $n$ distinct objects
There are $2$ ways for each object -
$1$. The object is chosen
$2$. The object is not chosen
$\therefore$ Number of ways =$2^n$
Multiplying each case gives us - $(p+1)(q+1)(r+1)(2^n)$ ways
Removing the case that no object is chosen gives us the required answer which gives us $(p+1)(q+1)(r+1)(2^n)-1$ ways