Question

Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

Answer 1

There are many ways to solve this problem. I will show two. I invite interested readers to present other answers, most probably more elegant ones.

(A) Solve by cases.
Because of the nature of the problem, the first digit can only begin with 6,7,8 or 9.
Case 6: 1 way
If it begins with a "6", there is only one way, namely consecutively, 6543210.
Case 7: 7 ways
If it begins with a 7, there is one way, consecutively, 7654321.
Since it finishes with a 1, a variation could be that one of the six following digits could have a gap with the previous, such as
7543210, or 7643210... There are 6 ways to do this.
total = 1+6 = 7 ways.
Case 8: 28 ways
consecutively=1 way
one digit skips 1 with previous digit=6 ways (e.g. 8654321...)
one digit skips 2 with previous digit=6 ways (e.g. 8543210,8743210...)
two digits skip 1 with previous digit=C(6,2) ways = 6*5/(2*1)=15 ways
total = 1+6+6+15 = 28 ways
Case 9: 64 ways
consecutively=1 way
one digit skips 1 with previous digit=6 ways
one digit skips 2 with previous digit=6 ways
one digit skips 3 with previous digit=6 ways
two digits skip 1 with previous digit=(6*5/(2*1))=15 ways
one digit skips 1 and one digit skips 2 with previous = 6*5=30 ways
total=1+6+6+6+15+30=64 ways

Grand total = 1+7+28+64 =120 ways.


(B) by computer program
The easiest way is to write a program and solve it to give a total of 120.
An example program could be:count:0;for x1:6 thru 9 do ( for x2:5 thru x1-1 do ( for x3:4 thru x2-1 do( for x4:3 thru x3-1 do( for x5:2 thru x4-1 do( for x6:1 thru x5-1 do( for x7:0 thru x6-1 do( count:count+1 ) ) ) ) ) ) )$print(count)$
Output shows 120 ways.