Total number of 4 digit numbers in which the digits are in descending order

Question

The question, as stated in the title, is:

Find the total number of 4 digit numbers in which the digits are in descending order.

Answer given: $\displaystyle {10 \choose 4} \text{ or } 210.$

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I've tried listing doing the possibilities of these numbers:

• Numbers starting with $9$:
  $9876,9875,...,9871,9870; 9765,9764,...9761,9760; 9654,9653,...,9651,9650;9543,9542,9541;9432,9431,9430,9420;9321,9320,... \text{and so on.}$

I could go on and count all the numbers in this way, but it's quite tedious and not very optimal. How do I use combinatorics here?

I would prefer a complete explanation.

EDIT: All the digits have to be different, as pointed out by @TonyK. I appologize for not clarifying earlier.

Answer 1

The answer has already been given, but I feel for explaining the theory behind the answer.

The solution is to find the amount of distinct subsets of the set of numbers from 0 to 9. Having these subsets, we can "reorder" the order in which the numbers appear (the order in sets the elements appear is not important). Therefore, computing the result is done by finding the number of unordered 4-combinations.

$$\frac{n!}{r! · (n - r)!} = \frac{10!}{4!(10 - 4)!} = 210$$

Answer 2

1. How many four-digit numbers are there in which all the digits are different?
2. Given four different digits, how many four-digit numbers can we make using them?
3. Divide your answer to 1 by your answer to 2.

Answer 3

Hint: Such a number is completely determined once you choose which four digits appear in it.

Answer 4

Observe that:

1) When you select a group of digits, there is only one way to arrange them in descending order.

2) Using any digit more than once in a sequence will violate the descending property.

3) To ensure descending order, $0$ may appear only as the least significant digit or none at all.

i. Sequences with $0$: There are ${9 \choose 3}$ of them, all ending with $0$.

ii. Sequences without $0$: There are ${9 \choose 4}$ of them.

Pascal's Rule gives us ${9 \choose 3} + {9 \choose 4} = {10 \choose 4} = 210$