bellcurve wrote:

If 6 different numbers are to be selected from integers 0 to 6, how many 6-digit even integers greater than 300,000 can be composed?

Anybody with a slot method?

--- Ans will be provided later.

Thanks,

BC

Why not..

Since the number must be greater than 300,000 :

first approach:

Take first number 3 or more than 3. Last number 0,2,4,6. For rest anything is fine

= 4*7*7*7*7*4

But this includes 300,000 as well. Since we want greater than 300,000, substract this 1 possiblity.

Ans = 4*7*7*7*7*4 -1

Second approach:

Take first number 3 or more than 3. For rest anything is fine

= 4*7*7*7*7*7

But we want only even numbers. Last digit can have 7 possible values out of which 4 are even.

Therefore multiply above number with 4/7

Total number of intergers equal to or more than 300000 = = 4*7*7*7*7*7 *4/7

But this includes 300,000 as well. Since we want greater than 300,000, substract this 1 possiblity.

Ans = 4*7*7*7*7*7*4/7 -1

Ans = 4*7*7*7*7*4 -1

Both approaches are slot method.

Hope it helps!

Edit: Sorry dint notice. we can use only digits 0-6 not 0-9. updated.

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