Internal telephone numbers in a university are composed of 5 digits. The first two digits can form any integer between (and including) 61 and 75, the third digit is an integer between 1 and 9 and the last two each take any integer value. Assume that it is disallowed to have a phone number with last three digits being 100, 101, or 102. In such case, the total number of possible internal telephone numbers that start with the digit 7 is.....
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is it 4500 - 15 = 4485
14+1 = 15 ways
1 to 9 = 9 ways (if we can count 1 and 9)
1000 ways - 3 = 997 ways for the ending
15*9*997 is what it looks like to me
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the first digit starts with a 7 mate, so its 5 possible choices times by 9 possible choices times by 100 possible choices for the last 2 digits including "00", since i have to avoid 100,101,102, i got to take of 15 possibilities so its 4500-15 = 4485
yeah, start with a 7 ... to many numbers to sort thru lol
70 71 72 73 74 75 = 6 ways then
6.9.997 is what I would see then
the last 3 digits can be from 000 to 999 (1000), excluding the 3 ways mentioned = 997 ways
61 to 75 = 15 ways, exclude the 6s and we get 6 ways to start
53,838 an option?
so its 5400 - 18 = 5382, its not multiple choice =\
wait, im still reading the question a little off. try this:
composed of 5 digits.
first two digits, starts with a 7: 7 __ 6 ways
3rd digit between 1 and 9: 9 ways
last two digits each take any integer value.
Assume that it is disallowed to have a phone number with last three digits being 100, 101, or 102
100 to 999 is where I was off at,
000 to 099 is 100 excluded ways, + the 3 mentioned total 103 excluded ways
897 ways total then sound right?
im still including that 3rd digit ....
00 to 99 is the last 2 digit options (100); exclude 00,01,02 (-3)
100-3 = 97 ways
6.9.97 = 5238