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 $1$.

  • $\begingroup$ What is $n$? You have not defined it. $\endgroup$ – N. F. Taussig Apr 23 '18 at 21:19
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    $\begingroup$ You have apparently ignored the or more part of "one or more." That being said, your equation is off a bit since you have an unnecessary $2^n-1$ appearing at the end. You didn't even define $n$ in this problem in the first place. The number of ways of selecting one or more items from $p$ identical, one or more from $q$ identical, and one or more from $r$ identical is $(p+1)(q+1)(r+1)$ and this can be seen by multiplication principle using the steps: "Choose how many copies of the first type of object do you take. Choose how many from the second type. Choose how many for the third" $\endgroup$ – JMoravitz Apr 23 '18 at 21:20
  • $\begingroup$ Having chosen two items of the first type is different than having chosen one item of the first type is further different than having chosen five of the first type etc... $\endgroup$ – JMoravitz Apr 23 '18 at 21:21
  • $\begingroup$ As an aside, $2^n-1$ is the number of ways to select one or more objects from $n$ distinct objects (which can also be seen by multiplication principle) $\endgroup$ – JMoravitz Apr 23 '18 at 21:31
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    $\begingroup$ Then you have either read the question incorrectly, copied the question incorrectly, or the book has an error. An answer of $(p+1)(q+1)(r+1)2^n-1$ would be an answer to the question of if you have $p$ identical items of a first type, $q$ identical of a second type, $r$ identical of a third type, and $n$ additional items all of which are distinct where you want to select items and must pick at least one of each of the first, second, and third types and may pick any number of the distinct items however you are not allowed to pick all of everything. $\endgroup$ – JMoravitz Apr 23 '18 at 21:53

The actual answer is $(p+1)(q+1)(r+1)(2^n)-1$ not $(p+1)(q+1)(r+1)(2^n-1)$

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


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'


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