Question

Number of ways of selecting one or more things from $p$ identical things of one type, $q$ identical things of one type and $r$ identical things of one type is given by

• A. $pqr$
• B. $(p+1)(q+1)(r+1)$
• C. $(p+1)(q+1)(r+1)-1$
• D. $pqr-1$

Answer 1

The correct option is C $(p+1)(q+1)(r+1)-1$

$p⟶$ identical of one type

$q⟶$ identical of one type

$r⟶$ identical of one type

Number of ways of selecting zero things or above form

(a) $p$ identical things =$p+1$

(b) $q$ identical things =$q+1$

(c) $r$ identical things =$r+1$

case for which all are zero $=1$

Total number of ways $=(p+1)(q+1)(r+1)-1$