Section B: Probability and Statistics [60 marks] Eleven cards each bears a single letter and together they can be made to spell the word COFFEEHOUSE. The 11 cards are arranged in a row. Three cards are selected from the eleven cards and the order of selection is not relevant. Find the number of possible selections that can be made

1. Identify the total number of letters and the frequencies of any repeated letters in the word 'COFFEEHOUSE'.

The word 'COFFEEHOUSE' contains $11$ letters in total.
The repeated letters are:

• 'O' appears $2$ times
• 'F' appears $2$ times
• 'E' appears $3$ times
  All other letters (C, H, U, S) appear once.

2. Calculate the number of possible arrangements by dividing the total permutations by the factorials of the repeated counts.

$\text{Number of arrangements} = \frac{11!}{2!2!3!}$

$= \frac{39916800}{2 \times 2 \times 6}$

$= 1663200$

1. Arrange the non-E letters, grouping the two F's together.

To ensure the two 'F's are together, we group them as a single block: (FF). We also temporarily set aside the three 'E's.
The letters to arrange are now the block (FF), C, O, O, H, U, S.
This gives a total of $7$ items, with the letter 'O' repeating $2$ times.
Number of ways to arrange these $7$ items $= \frac{7!}{2!} = 2520$.

2. Determine the number of ways to insert the three E's so that none are adjacent.

To ensure no two 'E's are next to each other, they must be placed in the spaces between or at the ends of the $7$ arranged items.
There are $8$ available spaces (represented by _):
_ (FF) _ C _ O _ O _ H _ U _ S _
Since the three 'E's are identical, we simply choose $3$ spaces out of the $8$.
Number of ways to place the 'E's $= \binom{8}{3} = 56$.

3. Calculate the total number of valid arrangements.

Total number of arrangements $= 2520 \times 56 = 141120$.

1. Identify the set of distinct letters available in the word COFFEEHOUSE.

The word COFFEEHOUSE contains the letters C, O, F, E, H, U, and S.
There are $7$ distinct letters.

2. Calculate the number of ways to select 3 different letters.

Since the order of selection is not relevant and identical letters are treated as indistinguishable, we simply choose $3$ letters from the $7$ distinct options available:
$\text{Number of possible selections} = \binom{7}{3} = 35$

1. Identify the frequencies of the letters to determine which can form a pair and the total number of distinct letters available.

The word COFFEEHOUSE consists of 11 letters with the following frequencies:

• E: 3
• F: 2
• O: 2
• C, H, U, S: 1 each
  There are 7 distinct letters in total. Only E, F, and O have a frequency of at least 2, so they are the only letters that can form a pair.

2. Calculate the number of selections with exactly one pair and one distinct letter.

To have exactly two cards with the same letter, we must choose 1 letter for the pair and 1 different letter for the single card.
Number of ways to choose the letter for the pair $= \binom{3}{1} = 3$ (from E, F, or O).
Number of ways to choose the single letter $= \binom{6}{1} = 6$ (from the remaining 6 distinct letters).
Total number of selections $= 3 \times 6 = 18$.