Question

if repetitions are not allowed, how many four digit numbers that a) are even b) greater than 7000 can be formed from the digits 1,2, ...,8,9

Answer 1

Case 1: the number starts with a 7 or a 9
There are
2 choices for the first digit,
5 choices for the last, (even) and
8 for the second and 7 for the third
for a total of 2*5*8*7=560 numbers
Case 2: the number starts with an 8
there is one choice for the first digit, 4 choices for the last, and
8 for the second and 7 for the third, for a total of
1*4*8*7=224 numbers

Sum of cases 1 & 2 = 560+224 = 784 even numbers between 7000 and 9999

Answer 2

oh the answer in the solution booklet says that it a) is 1344; any ideas how they got this?

Answer 3

Was the answer 1344 for even numbers between 7000 and 9999 without repetition?

Answer 4

oh it didn't say, so i'm not sure

Answer 5

My bad, the digits do not include zero, and the requirements are not combined for a and b. I`ll start over:
a. Even numbers between 7000 and 9999 without repetition
4 digits for the last
8 digits for the third
7 digits for the second
6 digits for the first
That makes 4*8*7*6=1344 numbers.

Answer 6

b. 4 digit numbers greater than 7000:
3 choices for the first digit (7,8 or 9)
8 choices for the second
7 choices for third and
6 digits for the fourth, for a total of
3*8*7*6 = 1008

Answer 7

OH I SEEE! HEHEHE THANK YOU

Answer 8

You're welcome!

Answer 9

would it be ok if you helped me with b) thanks mathmate!

Answer 10

Done! Check post following (a).

Answer 11

oh because the ans in the solution book is 1008

Answer 12

That's what I gave for (b). You may have missed the two posts send one after the other. The second one was sent "5 minutes ago".

Answer 13

ahhhhhhh MY APOLOGIES!!!!!!!!!!!!!!!!!

Answer 14

No problem at all! :)

Answer 15

@mathmate, are you sure you are counting even numbers between 7000 and 9999 without repetition in as:

>>4 digits for the last
>>8 digits for the third
>>7 digits for the second
>>6 digits for the first
>>That makes 4*8*7*6=1344 numbers.

Answer 16

Not really, what I got was
"Sum of cases 1 & 2 = 560+224 = 784 even numbers between 7000 and 9999"
was not what the question asked for.

It was divided into two cases where the number starts with 7 or 9 (case 1) or starts with 8 (case 2).