sanju09 wrote:Lola forgot last three digits of one telephone number. All she remembers is that the missing digits were distinct, non-zero, and were in ascending order. Based on this information, she dials one number. What is the probability that she dials a wrong number?
[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]
Some examples of 3-digit numbers that meet the above criteria include 123, 156, 246, 489, 579, etc.
So, the question requires us to determine how many such numbers are possible.
Approach #1: Take the 9-digit number 123456789 and choose any 6 digits to eliminate.
Example: if we eliminate the digits 1,3,4,5,6 and 8, we're left with the 3-digit number 279
In how many ways can we select 6 digits to eliminate? Well, there are 9 digits altogether, and we must select 6. This can be accomplished in 9C6 ways (or 84 ways)
Approach #2: Take the 9 digits (1,2,3,4,5,6,7,8 and 9) and choose any 3.
Now list those 3 digits in ascending order, and you have a 3-digit number that meets the given criteria.
Note: there's only 1 way to list 3 unique digits in ascending order.
In how many ways can we select 3 digits? Well, there are 9 digits altogether, and we must select 3. This can be accomplished in 9C3 ways (or 84 ways)
Aside: Notice that 9C6 = 9C3 = 84. In general, nCr = nC(n-r)
Okay, so we now know that there are 84 different 3-digit numbers, and only 1 of those numbers is the correct number. In other words, 83 of the 84 numbers are the wrong number.
So . . . P(wrong number) = 83/84
Cheers,
Brent
