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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.

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Hint: If you pick any four distinct numbers from the set $\{0,1,2,3,4,5,6,7,8,9\}$, there is one and only one way to arrange them so that they are in decreasing order.

For the numbers whose digits are in increasing order, simply observe that the leftmost (most significant, leading) digit cannot be zero.

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    $\begingroup$ Right, I see now, this makes sense. Obviously my problem was not being able to see the simple fact that there is only one way to arrange the number in decreasing order. Thanks! $\endgroup$ – rheotron Mar 30 '15 at 5:30
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Shouldn't the answer be 9C4 + 9C3 for question a and 9C4 for question b ?

Case 1. The first term cannot be zero of a number of format "abcd".

So we have 9 choices.

9C4 for either way of arrangement.

Case 2. Only the last term d can be 0. So we have 9 choices for a,b,c. 9C3 ways of arranging in strict decreasing order.

So when we have strict increasing from left to right we have 2 cases the last term being zero and none of the terms being zero. 9C3 + 9C4

When we have strict decreasing from left to right we have only 1 cases as the last term and first term can never be zero. 9C4.

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(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.

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