Question

You pick 6 digits (0-9) at random without replacement, and write them in the order picked. What is the probability that you have written the first 6 digits of your phone number? Assume there are no repeats of digits in your phone number.

Answer 1

Answer:

1/151200

Step-by-step explanation:

The number of ways to pick an exact order of any 6 different numbers from 10

$_{6}P^{10}\textrm{}$

=$\frac{10!}{4!}$

= 10×9×8×7×6×5 = 151200 ways

which means the first one can be any of 10 numbers

The second one can be any of 9 numbers

The 3rd one can be any of 8 numbers and so on

Now the 6th one can be any of 5 numbers

Thus there are  10×9×8×7×6×5 = 151200 ways  to choose the number

so the required probability is 1/151200