Question

A school labels its time capsules using 3 distinct uppercase letters from A to Z. Each label must satisfy the following conditions:

• The letter T must appear somewhere in the label.
• The three letters must be in strict alphabetical order.

How many such distinct labels are possible?

• A. 350
• B. 300
• C. 220
• D. 240

Answer 1

Answer: (B)

300

Answer 2

Let the three distinct uppercase letters be $L_1, L_2, L_3$. The letters must be in strict alphabetical order, so $L_1 < L_2 < L_3$. The letters are chosen from the 26 uppercase letters A to Z. The letter T must appear somewhere in the label. T is the 20th letter of the alphabet.

Let's consider the possible positions of the letter T in the alphabetically ordered label:

Case 1: T is the first letter ($L_1 = T$).

If $L_1 = T$, then the letters must satisfy $T < L_2 < L_3$. $L_2$ and $L_3$ must be distinct letters chosen from the letters that come after T in the alphabet. The letters after T are U, V, W, X, Y, Z. There are 6 such letters. We need to choose 2 distinct letters from these 6 letters for $L_2$ and $L_3$. Once chosen, their order is fixed by the alphabetical requirement. The number of ways to choose 2 letters from 6 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Number of labels in this case = $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$.

Case 2: T is the second letter ($L_2 = T$).

If $L_2 = T$, then the letters must satisfy $L_1 < T < L_3$. $L_1$ must be a letter that comes before T. The letters before T are A, B, ..., S. There are 19 such letters. $L_3$ must be a letter that comes after T. The letters after T are U, V, ..., Z. There are 6 such letters. We need to choose 1 letter for $L_1$ from the 19 letters before T and 1 letter for $L_3$ from the 6 letters after T. The number of ways to choose $L_1$ is $\binom{19}{1} = 19$. The number of ways to choose $L_3$ is $\binom{6}{1} = 6$. Since the choice of $L_1$ is independent of the choice of $L_3$, the total number of labels in this case is the product of the number of ways to choose $L_1$ and $L_3$. Number of labels in this case = $\binom{19}{1} \times \binom{6}{1} = 19 \times 6 = 114$.

Case 3: T is the third letter ($L_3 = T$).

If $L_3 = T$, then the letters must satisfy $L_1 < L_2 < T$. $L_1$ and $L_2$ must be distinct letters chosen from the letters that come before T in the alphabet. The letters before T are A, B, ..., S. There are 19 such letters. We need to choose 2 distinct letters from these 19 letters for $L_1$ and $L_2$. Once chosen, their order is fixed by the alphabetical requirement. The number of ways to choose 2 letters from 19 is given by the combination formula. Number of labels in this case = $\binom{19}{2} = \frac{19 \times 18}{2 \times 1} = 19 \times 9 = 171$.

The three cases cover all possible labels that contain T and are in strict alphabetical order. These cases are mutually exclusive. The total number of distinct labels possible is the sum of the numbers from the three cases. Total number of labels = (Number of labels with T as $L_1$) + (Number of labels with T as $L_2$) + (Number of labels with T as $L_3$) Total number of labels = $15 + 114 + 171 = 300$.

Alternatively, we can calculate the total number of distinct labels with 3 distinct uppercase letters in alphabetical order and subtract the number of such labels that do not contain T. The total number of ways to choose 3 distinct letters from 26 letters and arrange them in alphabetical order is $\binom{26}{3}$. $\binom{26}{3} = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} = 26 \times 25 \times 4 = 2600$. The number of distinct labels with 3 distinct uppercase letters in alphabetical order that do not contain T are those formed by choosing 3 letters from the 25 letters excluding T. The number of ways to choose 3 letters from 25 is $\binom{25}{3}$. $\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 100 \times 23 = 2300$. The number of labels containing T is the total number of labels minus the number of labels not containing T. Number of labels containing T = $\binom{26}{3} - \binom{25}{3} = 2600 - 2300 = 300$.

Both methods give the same result.