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 and $n$ different things is given by

• A. $pqrn-1$
• B. $2^{n}pqr-1$
• C. $(p+1)(q+1)(r+1)2^n-1$
• D. none of these

Answer 1

The correct option is C $(p+1)(q+1)(r+1)2^n-1$

$p⟶$one type

$q⟶$one type

$r⟶$one type

$n⟶$diffrent

Number of selecting zero or more things of

(a) $p$ idntical things $p+1$

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

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

Number of ways of selecting $n$ diffrent things $={\left(2\right)}^n$

Number of ways of selecting zero things $=1$

Total number of ways $=(p+1)(q+1)(r+1)2^n-1$