How many six digit even numbers are possible if the leftmost digit cannot be zero

450000 six digit even numbers are possible if the leftmost digit cannot be zero.

Step-by-step explanation:

We have to find Total six digit even numbers.

To form an even of 6 digit we have 10 digits to place in first 5 position starting from left most, here repetition is allowed so, No ways of doing this is  and number of ways selecting one digit from 5 digit for ones place is [/tex]^{5}\textrm{P}_{1}=\farc{5!}{(5-1)!}=\frac{5!}{4!}=5[/tex]

So, Total even numbers =

This includes number which has zero on the left most place.

To get these numbers we fir 0 at place 2 left 4 places has 9 options and ones place has 5 options as before.

So, Number of such cases =

Thus, total Number of 6 Digit number even number = 500000 - 50000 = 450000

Therefore, 450000 six digit even numbers are possible if the leftmost digit cannot be zero.

its 3

step-by-step explanation:

because i like 3

i cant see the graph

step-by-step explanation: