Number of possible integers less than 100000 such that the digits 4,5,6 appear in that order I have been stuck on this problem for some time and I am not sure how to approach it:
"How many positive integers less than 100,000 have digits containing 4,5,6 in that particular order?" (by this it means that the digits "456" must appear one after another in the number)
I am thinking that I would multiply 10X10X3X2X1 since the two of the five digit spaces can hold any digit between 0-9 and the other 3 must have 456 however I do not think this considers the order of the digits or the fact that the smallest possible integer would be 456. (if that makes any sense)
How would I modify this so that the order of the numbers matter?
What would the answer be to this?
thanks in advance!
 A: Reading the problem statement, I will think that digits $4,5,6$ appear only once, do not need to be consecutive digits, and the sequence is always $4$ first and $6$ last.
From $5$ digits, we choose $3$ to put $4,5,6$ in that order. $\binom{5}{3}$ ways.
The remaining $2$ digits can be anything but $4,5,6$. $7^{2}$ ways.
So there are $\binom{5}{3}7^{2}$ possible integers
A: There are a total of $5$ digits
The $3$ digit block $4,5,6$ can be placed in $3$ positions. ($1,2,3$ and $2,3,4$ and $3,4,5$)
$3×10×10$
So there are a total of $300$ numbers in which the digits $4,5,6$ appear in this order consecutively
A: We can represent any nonnegative integer less than $100000$ as a five-digit string by appending leading zeros as necessary.  For instance, we can represent $456$ as $00456$ and $4956$ as $04956$.
There are ten possible positions in which the first $4$, first $5$, and first $6$ could appear:
$456\square\square$
$45\square 6 \square$
$45 \square \square 6$
$4 \square 56 \square$
$4 \square 5 \square 6$
$4 \square \square 56$
$\square 456 \square$
$\square 45 \square 6$
$\square 4 \square 56$
$\square \square 456$
Any position before the first $4$ can be filled in $7$ ways since it cannot be a $4$, $5$, or $6$.
Any position after the first $4$ but before the first $5$ can be filled in $8$ ways since it cannot be a $5$ or $6$.
Any position after the first $5$ but before the first $6$ can be filled in $9$ ways since it cannot be a $6$.
Any position after the first $6$ can be filled in $10$ ways.
Thus, the number of nonnegative integers less than $100000$ in which the digits $4$, $5$, and $6$ appear in that order is 
$$1 \cdot 1 \cdot 1 \cdot 10 \cdot 10 + 1 \cdot 1 \cdot 9 \cdot 1 \cdot 10 + 1 \cdot 1 \cdot 9 \cdot 9 \cdot 1 + 1 \cdot 8 \cdot 1 \cdot 1 \cdot 10 + 1 \cdot 8 \cdot 1 \cdot 9 \cdot 1 + 1 \cdot 8 \cdot 8 \cdot 1 \cdot 1 + 7 \cdot 1 \cdot 1 \cdot 1 \cdot 10 + 7 \cdot 1 \cdot 1 \cdot 9 \cdot 1 + 7 \cdot 1 \cdot 8 \cdot 1 \cdot 1 + 7 \cdot 7 \cdot 1 \cdot 1 \cdot 1$$
If you make the assumption that the digits $4$, $5$, and $6$ each appear exactly once, you obtain Rezha Adrian Tanuharja's answer.  This is because there would be only seven choices for each of the other two positions.
