How many distinct even five digit numbers greater than 60,000 exist?

Question

How many distinct even five digit numbers greater than 60,000 exist if each of the digits 3,4,5,6 and 9 must be used exactly once in each even number?

Answer 1

Let's start with the last digit, which must be 4 or 6.

If 4, then you have 2 choices (6 or 9) for the first digit. The remaining digits have 3 choices, 2 choices and 1 choice respectively.

That works out to 2 x 3 x 2 x 1 x 1 = 12 ways

But if you use 6 as the last digit, then you only have 1 choice (9) for the first digit and 3 choices, 2 choices, 1 choice for the middle digits.

1 x 3 x 2 x 1 x 1 = 6 ways.

Added together there are 18 ways. Here they are enumerated:

63594
63954
65394
65934
69354
69534

93456
93546
93564
93654
94356
94536

95346
95364
95436
95634
96354
96534

Answer 2

The questions asks about EVEN numbers:

this means it has to end in either the 6 or the 4
as it is above 60000 it have to start with either a 6 or a 9
options are therefore
6xxx4 ... 63594, 63954, 65394, 65934, 69354, 39534
9xxx4 ... 6 options as above
9xxx6 ... 6 options as above

6 +6 +6 = 18

ANS: 18 distinct 5 digit numbers greater than 60,000 exist if each of the digits 3,4,5,6 and 9 must be used exactly once in each EVEN number

Answer 3

the first digit must be 6 or 9 to make it > 60000
the last digit must be 6 or 4 to make it even
so the possible pairs of first and last digit are 64, 94, 96
there are 3! = 6 ways to arrange the other 3 digits to complete the number.

asnwer = 18

Answer 4

2*4*3*2*1=48