How many 5-digit numbers exist between 65,000 and 69,999 if no digit is to be repeated in each number?
on the first digit you can use 5-9 so you have 4 options on the other digits you can use any digit between 0-9, 10 options So without repeating the first digit you are only left with 9 options for the second, 8 options for the third digit and 7 options on the last digit Answer is 4 x 9 x 8 x 7 = 2016
cool. thanks! :)
oh. it's rather 5 x 8 x 7 x 6. because at the time you choose a digit in the 3rd place, you have already used 2 digits (6 and the second digit).
oh no. you were right for the '4', but im right for 8x7x6.
so what is the answer? haha. im confused
number: a b c d e a is '6' (no choice) b is in 5, 7, 8, 9 (4 choices) c is in {0,1,....,9} \ {a,b} (8 choices) d is in {0,1,...,9} \ {a,b,c} (7 choices) e is ..... (6 choices) -> \(4\times8\times7\times6\).
ok, i gotchya! thanks! :)
nope, since you can't use 6. and you can't use the second digit. -> you used up two digits already.
... you can't use 6, so the choice for the second digit of the number is limited to 5, 7, 8, 9, not 6! -> only 4 choices.
His answer is right. I cannot use 6 on any digit.
:) few! yea these questions are a mess
yeah. stupid math project. so the answer is...? im confused again.
4 x 8 x 7 x 6. ——— number: a b c d e a is '6' (no choice) b is in 5, 7, 8, 9 (4 choices) c is in {0,1,....,9} \ {a,b} (8 choices) d is in {0,1,...,9} \ {a,b,c} (7 choices) e is ..... (6 choices) ———
ok, thank you!
1344 i guess.
good enough.
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