CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

How many three-digit numbers that are divisible by 5 can be formed using the digits 0, 2, 3, 5, 7, if no digit occurs more than once in each number


A
21
loader
B
20
loader
C
18
loader
D
15
loader

Solution

The correct option is A 21
Since each desired number is divisible by $$ 5 $$, so we must have $$ 0\  or   \ 5 $$ at the unit place.
Case 1, when $$ 0 $$ is at the units place
Then the hundreds and tens place can be filled up by remaining $$ 4 $$ digits in $$ \dfrac { 4! }{ (4-2)! } =\quad \dfrac { 4! }{ 2! } =\quad 4\times 3=\quad 12 $$ ways

Case 2, when $$ 5 $$ is at the units place
Then the ten's digit can be filled by the digits $$ 0,2,3,7 $$ in $$ 4 $$ ways
And the $$100's$$ digit can be filled by any of the remaining $$ 2 $$ digits in $$ 2 $$ ways as $$ 0 $$ cannot take Hundred's place, then number wont be a $$ 3 $$-digit number then.
So, total number ways $$ = 12 + 4 + 2 = 18 $$ ways

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image