Seven cards each bearing a letter , can be arranged to spell the word 'DOUBLES' .How many three - letter code- words can be formed from these cards ?

Answer

To solve the problem of how many three-letter code words can be formed from the letters of the word "DOUBLES", we can follow these steps:

### Step 1: Identify the total number of letters
The word "DOUBLES" consists of 7 letters: D, O, U, B, L, E, S. None of these letters are repeated.

### Step 2: Select 3 letters from the 7 letters
We need to select 3 letters from the 7 available letters. The number of ways to choose 3 letters from 7 is given by the combination formula:
\[ \text{Number of ways to choose 3 letters} = \binom{7}{3} \]

### Step 3: Calculate the combination
Using the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
we can calculate:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} \]
This simplifies to:
\[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 4: Arrange the selected letters
After selecting 3 letters, we can arrange these 3 letters in different ways. The number of arrangements of 3 letters is given by:
\[ 3! = 6 \]

### Step 5: Calculate the total number of code words
Now, we multiply the number of ways to choose the letters by the number of arrangements:
\[ \text{Total code words} = \binom{7}{3} \times 3! = 35 \times 6 = 210 \]

Thus, the total number of three-letter code words that can be formed from the letters of "DOUBLES" is **210**.

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