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How many 5 digit telephone numbers are there that do not have '75' in that order in any place in the number? (assume a telephone number can start with 0)

2007-10-23 06:59:47 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

First, let's find all the possible 5-digit numbers:

10 * 10 * 10 * 10 * 10 = 100,000 possible numbers

Now, let's find how many 5-digit numbers have '75' in them, in that same order:

* The '+' symbol will represent a variable that can be any number

1) 75+++ 10 * 10 * 10 = 1000
2) +75++ 10 * 10 * 10 = 1000
3) ++75+ 10 * 10 * 10 = 1000
4) +++75 10 * 10 * 10 = 1000

There are only four thousand 5-digit numbers that have '75' in that order.

Now, subtract the two numbers:

100,000 - 4,000 = 96,000

* The math teacher below me is forgetting what the question-asker said: "assume a telephone number can start with 0," in which case it is can be 10*10*10*10*10, and not 9*10*10*10*10. Also, this would make the second part the the answer wrong too.

2007-10-23 07:04:23 · answer #1 · answered by عبد الله (ドラゴン) 5 · 1 4

We can have 10^5 (100000) telephone numbers, Total.
Now, I assume we want to eliminate the ones that have '75' next to each other, any place in the number. The telephone number basically looks like this: x x x x x and x can be any one of the ten digits.
The '75' can be in one of 4 positions in the telephone number:
7 5 x x x
x 7 5 x x
x x 7 5 x
x x x 7 5

For each of the 4 positions, we have 10^3 possibilities for the other numbers (the x's in the diagram).
The quantity of 5-digit telephone numbers that do not have '75' in that order are 10^5 - 4 * 10^3.

2007-10-23 07:10:03 · answer #2 · answered by Hiker 4 · 1 1

People are not taking into account that the phone # cannot have a 0 for the first digit.

1st number: 9 choices
2nd number: 10 choices
3rd number: 10 choices
4th number: 10 choices
5th number: 10 choices

9 * 10 * 10 * 10 * 10 = 90,000

But you still have to take into account the combinations of those 90,000 that you would have "75" right in a row.

75 _ _ _ 10 * 10 * 10 = 1,000 combinations
_ 75 _ _ 9 * 10 * 10 = 900 combinations
_ _ 75 _ 9 * 10 * 10 = 900 combinations
_ _ _ 75 9 * 10 * 10 = 900 combinations

3,700 combinations that include "75" right in a row.

90,000 total combinations - 3,700 specific combinations excluded = 86,300 combinations

2007-10-23 07:12:26 · answer #3 · answered by Bryana B 2 · 0 2

Hiker is right. Bryanna, she clearly stated in the question:
assume that the telephone number CAN start with 0.

2007-10-23 07:24:17 · answer #4 · answered by swd 6 · 0 0

You have 10 digits and 5 slots
So you hav 10 times 10 times 10 times 10 times 10
Then have to subtract 4 for 75XXX ,X75XX,XX75X,XXX75

2007-10-23 07:08:31 · answer #5 · answered by lillilou 7 · 0 2

99999 + the all zero number - 4 cuz there are 4 positions the 75 can be in right?
so that would be..... um about 99996

2007-10-23 07:14:53 · answer #6 · answered by dire_st 3 · 0 2

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