Question

The number of positive integers which can be formed by using any number of digits from $0,1,2,3,4,5$but using each digit not more than once in each number, is equal to:

• A. $1200$
• B. $1500$
• C. $1600$
• D. $1630$

Answer 1

The correct option is D

$1630$

Explanation for the correct option.

Finding the number of positive integers:

Given numbers: $0,1,2,3,4,5$

Number of given digits are $6$.

Case (i): Single digit number from $0-5$ but 0 is neither positive nor negative number then we can be written in ${}^5{P}_1\text{ways =}5$

Case (ii): Two-digit numbers from $0-5$ digit can be done in,

$\begin{array}{rcl}{}^6{P}_2–{}^5{P}_1& =& 30–5\\ & =& 25\end{array}$

Case (iii): Three-digit numbers from $0-5$ digit can be done in,

$\begin{array}{rcl}& \Rightarrow & {}^6{P}_3{–}^5{P}_2=\frac{6!}{3!}-\frac{5!}{3!}\\ & =& 120–20\\ & =& 100\end{array}$

Case(iv): Four-digit numbers from $0-5$ digit can be done in,

$\begin{array}{rcl}& \Rightarrow & {}^6{P}_4{–}^5{P}_4=\frac{6!}{2!}-\frac{5!}{2!}\\ & =& \frac{720}{2}–\frac{120}{2}\\ & =& 360-60\\ & =& 300\end{array}$

Case(v): Five-digit numbers from $0-5$ digit can be done in,

$\begin{array}{rcl}& \Rightarrow & {}^6{P}_5{–}^5{P}_4=6!-5!\\ & =& 720-120\\ & =& 600\end{array}$

Case(vi): Six-digit numbers from $0-5$ digit can be done in,

$\begin{array}{rcl}& \Rightarrow & {}^6{P}_6{–}^5{P}_5=6!-5!\\ & =& 720-120\\ & =& 600\end{array}$

Adding all the numbers together, we get;

$5+25+100+300+600+600=1630$

Therefore, the total number greater than 3000 will be 1630.

Hence, option (D) is the correct answer.