Question

[
begin{array}{l}
cdot n !=n(n-1)(n-2)(n-3) ldots
ldots(3)(2)(1)
end{array}
]
Numbers using 1,2,3,4,5
No. of numbers ( =5 !=120 )
Numbers using 0,1,2,4,5
No. of numbers ( =5 !-4 !=120-24=96 )
There are 4! numbers beginning with 0
( therefore ) Total number of numbers ( =120+96=216 )

# 3 (101.) A five digit number divisible by 3 is to be formed using the numbers 0,1,3,4 well and 5 without repetition. The total number of ways this can be done is D 33+ (a) 216 Net 240 (b) 600 (d) 3125 ca 100 10-m min 33 100-mami

Solution