Number of strictly increasing and decreasing 4 digit numbers? I have two questions:
(a) Count the 4 digit numbers whose digits decrease strictly from left to right and;
(b) Count the 4 digit numbers whose digits increase strictly from left to right
I have the answers which are $10 \choose 4$ and $9 \choose 4$ respectively. Can someone explain how to come to these answer? The number of ways to choose 4 elements of 10 or 9 respectively seems to simplistic.

(a) Since the digits must strictly decrease from left to right, there are 10C4 such numbers. For any selection of four digits from the ten digits, there is exactly one way of arranging those four digits in a strictly decreasing order from left to right. We need not worry about leading zeroes because the order is decreasing, and 0 can never be the left most digit. Hence 10C4.
(b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The answer is therefore, 9C4.
