Determine how many strings can be formed by ordering the letters ABCDE where A appears before C and C appears before E. I need help solving this problem:
Determine how many strings can be formed by ordering the letters ABCDE where A appears before C and C appears before E.
My teacher took the "intersection" of this and it's confusing me - what does she mean by intersection? She also treated this as a combination problem but how is that possible if order matters since a change of letters will result in a new string?
Any tips would be appreciated.
 A: There are several ways to solve this problem, and I can’t tell from your description which one your teacher actually used. Here are two of them, and the second one does use combinations.
Since $A$ must appear before $C$ and $C$ before $E$, you have a skeleton _A_C_E_: the $B$ and $D$ must go into the underlined slots. Then can go into two different slots in $4\cdot 3=12$ different ways: there are $4$ slots into which the $B$ can go, and then the $D$ can go into any of the $3$ remaining slots. They can also go into the same slot: in that case there are $4$ choices for the slot and $2$ orders for $B$ and $E$ in that slot, so there are $4\cdot 2=8$ possibilities. Altogether, then, there are $12+8=20$ possible arrangements of the letters.
Or you can pretend for a moment that $B$ and $D$ have each been replaced by $X$. Then we have a string of $5$ letters, $A,B,C,X$, and $X$, and if we know where the two $X$s are, we know the whole string, because the $A,C$, and $E$ have to appear in that order. For instance, if we have __XX_, the string must be $ACXXE$. Thus, there is one such string for every way of choosing $2$ of the $5$ slots and filling them with the two $X$s. We can choose $2$ of the $5$ slots in $\binom52=10$ ways, so there are $10$ such strings. Finally, each of those strings gives us $2$ strings of $A,B,C,D$, and $E$ with $A$ before $C$ and $C$ before $E$, because we can replace the first $X$ with $B$ and the second with $D$, or we can replace the first with $D$ and the second with $B$. For instance, $ACXXE$ gives us $ACBDE$ and $ACDBE$. Thus, there are $10\cdot2=20$ strings of the kind that we want.
A: The intersection is just the set-theoretical way to say "and". Notice that A<C<E is the same as A<C and C<E.
Hint: Let's say that you have your letters in a bag and that you want to put them together using this notion of ordering. So, the first thing you would do is try to make sure that they are respecting the ordering, in other words, you would like to choose three indices $i<j<k$ such that you are gonna place A, C and E. In how many ways can you do this? Well $\binom{5}{3}.$ Notice that the first index has to be occupied by A, the second one by C and the third one by E. So you are left with 2 indices in which you have to place B and D. So you have $2$ possible orderings. Use the product rule and you are good to go!
A: Method 1:  We have five positions to fill.  There are five ways to place the B, which leaves four ways to place the D.  Once those positions have been filled, the remaining positions must be filled with A, C, and E in that order.  Hence, there are $5 \cdot 4 = 20$ admissible arrangements.
Method 2:  We have five positions to fill.  We choose three positions for the letters A, C, and E, which can be done in $\binom{5}{3}$ ways.  Since A must appear before C and C must appear before E, there is only one way to arrange A, C, and E in these positions.  There are $2!$ ways to arrange B and D in the remaining two positions, so there are
$$\binom{5}{3}2!$$
admissible arrangements.
Method 3:  There are $5!$ ways to arrange five distinct letters.  Within a given arrangement, there are $3!$ ways to arrange the letters A, C, and E while holding the positions of the other letters fixed.  Only one of these $3!$ arrangements places A before C and C before E.  Thus, by symmetry, there are
$$\frac{1}{3!} \cdot 5! = 20$$
admissible arrangements.
