<s>[INST] What is the probability that a four-digit number formed by randomly selecting four-digits without replacement is greater than $4321$? A four-digit number is formed by randomly selecting four digits, without replacement, from the set $D = \{ 1, 2, 3, 4, 5, 6, 7\}$. What is the probability that the resulting number is greater than $4321$?
Attempt: The odds of choosing a $5$, $6$, or $7$ are $1/7 + 1/7 + 1/7 = 3/7$.  Then the odds of choosing a $4$ for the first number are $1/7$.  If my second number is $5$, $6$, or a $7$, I know the number will be greater than $4321$. This is $3/6$ so $1/7 \cdot 3/6 = 3/42$.  Then I'm getting stuck.
 [/INST] What you have done thus far is correct.  That leaves cases in which the first number is $4$ and the second number is $3$.  The probability of first obtaining $4$, then obtaining $3$ is $\frac{1}{7} \cdot \frac{1}{6} = \frac{1}{42}$.  We now consider two cases: the third digit is larger than $2$ or the third digit is $2$ and the fourth digit is larger than $1$.  
The first digit is $4$, the second digit is $3$, and the third digit is larger than $2$:  The remaining digits are $1, 2, 5, 6, 7$. Of these, three are larger than $2$. Hence, the probability that the first digit is $4$, the second digit is $3$, and the third digit is larger than $2$ is 
$$\frac{1}{7} \cdot \frac{1}{6} \cdot \frac{3}{5}$$
The first digit is $4$, the second digit is $3$, the third digit is $2$, and the fourth digit is larger than $1$:  Once $4$, $3$, and $2$ have been selected, the remaining digits are $1, 5, 6, 7$, of which three are larger than $1$.  Hence, the probability that the first digit is $4$, the second digit is $3$, the third digit is $2$, and the fourth digit is larger than $1$ is 
$$\frac{1}{7} \cdot \frac{1}{6} \cdot \frac{1}{5} \cdot \frac{3}{4}$$
Hence, the probability of randomly selecting a four-digit number without replacement from the set $D = \{1, 2, 3, 4, 5, 6, 7\}$ that is larger than $4321$ is 
$$\frac{3}{7} + \frac{1}{7} \cdot \frac{3}{6} + \frac{1}{7} \cdot \frac{1}{6} \cdot \frac{3}{5} + \frac{1}{7} \cdot \frac{1}{6} \cdot \frac{1}{5} \cdot \frac{3}{4}$$
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