[u]Set 4[/u] 


[b]p10.[/b] Eve has nine letter tiles: three $C$’s, three $M$’s, and three $W$’s. If she arranges them in a random order, what is the probability that the string “$CMWMC$” appears somewhere in the arrangement?


[b]p11.[/b] Bethany’s Batteries sells two kinds of batteries: $C$ batteries for $\$4$ per package, and $D$ batteries for $\$7$ per package. After a busy day, Bethany looks at her ledger and sees that every customer that day spent exactly $\$2021$, and no two of them purchased the same quantities of both types of battery. Bethany also notes that if any other customer had come, at least one of these two conditions would’ve had to fail. How many packages of batteries did Bethany sell?


[b]p12.[/b] A deck of cards consists of $30$ cards labeled with the integers $1$ to $30$, inclusive. The cards numbered $1$ through $15$ are purple, and the cards numbered $16$ through $30$ are green. Lilith has an expansion pack to the deck that contains six indistinguishable copies of a green card labeled with the number $32$. Lilith wants to pick from the expanded deck a hand of two cards such that at least one card is green. Find the number of distinguishable hands Lilith can make with this deck.


PS. You should use hide for answers.
To solve the problem, we need to determine the probability that the string "CMWMC" appears somewhere in the arrangement of nine letter tiles consisting of three C's, three M's, and three W's.

1. **Calculate the total number of arrangements of the nine letter tiles:**
   The total number of ways to arrange the nine tiles (three C's, three M's, and three W's) is given by the multinomial coefficient:
   \[
   \frac{9!}{3!3!3!}
   \]
   Calculating this, we get:
   \[
   \frac{9!}{3!3!3!} = \frac{362880}{6 \cdot 6 \cdot 6} = \frac{362880}{216} = 1680
   \]

2. **Calculate the number of favorable arrangements where "CMWMC" appears as a block:**
   If "CMWMC" appears as a block, we can treat this block as a single entity, denoted by \(X\). This reduces the problem to arranging the block \(X\) along with the remaining four letters (one C, one M, and two W's).

   The number of ways to arrange these five entities (X, C, M, W, W) is given by:
   \[
   \frac{5!}{1!1!2!1!}
   \]
   Calculating this, we get:
   \[
   \frac{5!}{1!1!2!1!} = \frac{120}{2} = 60
   \]

3. **Calculate the probability:**
   The probability that "CMWMC" appears as a block in the arrangement is the ratio of the number of favorable arrangements to the total number of arrangements:
   \[
   \frac{60}{1680} = \frac{1}{28}
   \]

The final answer is \(\boxed{\frac{1}{28}}\).