Probabilityyyyy?

if the first digit of a seven-digit telephone number can never be a zero, what is the probability that your telephone number will have increasing digits (such as 234-5678).




please guys. i have a million of these for homework and i am getting the biggest head ache from them. i am doing so good in every subject but this one && it is frustrating. so in order to get the other 10 problems that I have on these could you tell me step-by-step on how you got the answer. i already tried it && ended up getting it wrong & becoming confused. so please solve this one for me so i can use this as an example.



thank youuuu

OK -
If your telephone number has seven digits, and the first cannot be zero, then the total number of telephone numbers is 9*10*10*10*10*10*10 or 9,000,000. (Finding the total number of possibilites is always the first step of these problems.)


If you think about the pattern, there are only three possibilities -
a) the number starts with 1.
If the number starts with 1, we can have up to two "breaks" in sequence - as in 124 5689 or 134 6789. Mathematically, this is 8 things taken 6 at a time for the rest of the numbers, but order matters, so this is 8!/6!*2!*, or 28 possibilities.
Here are all those possibilities:
Start with the digit one and remove all these combinations -
23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89.

b) the number starts with 2 - then we can only have one "break" in the numbers, and there are only 7 possibilites [you can't leave 2 out] - or, mathematically, seven things taken 6 at a time, or 7!/6!, which equals 7 -
2345678
2345679
2345689
2345789
2346789
2356789
2456789

c) the number starts with 3 - there is only one of these, 345 6789.

This gives us a total of 28+7+1 = 36 different combinations. So the chance of this happening is 36/9000000 or 0.0004%.