Question

The number of four digit numbers that can be formed with 0, 1, 2, 3, 4, 5 is:

• A. ${}^6{P}_4$
• B. 5.${}^6{P}_3$
• C. ${}^6{P}_4-{}^5{P}_4$
• D. ${}^6{P}_4-{}^5{P}_3$

Answer 1

The correct option is D ${}^6{P}_4-{}^5{P}_3$

No. of digits = 6.
$\to$ The no. of permutations of 6 objects in 4 places = ${}^6{P}_4$.
$\to$ However, we need to subtract the case where the no. started with 0. i.e., $0543$ = ${}^5{P}_3$

So, the total no. of ways = ${}^6{P}_4{-}^5{P}_3$

Ans = ${}^6{P}_4{-}^5{P}_3$