How many integers between 1000 and 9999 (4 digits), contain only the digits 1,2,3,4,5,6 and 7, and no digit appears twice?

1. Break Down the Problem

We need to find how many 4-digit integers can be formed using the digits {1, 2, 3, 4, 5, 6, 7}, where no digits can repeat.

2. Relevant Concepts

For a 4-digit number, we need to select and arrange 4 digits from a set of 7. The number of ways to do this is given by the permutation formula:

$P(n, r) = \frac{n!}{(n-r)!}$

where:

• $n$ is the total number of digits available,
• $r$ is the number of digits to choose (which is 4 in this case).

3. Analysis and Detail

Here, $n = 7$ (the digits are {1, 2, 3, 4, 5, 6, 7}) and $r = 4$.

Using the permutation formula: $P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!}$

Calculating $7!$ and $3!$:

• $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$
• $3! = 3 \times 2 \times 1 = 6$

Now, substituting into the formula: $P(7, 4) = \frac{5040}{6} = 840$

4. Verify and Summarize

We find that the total number of ways to arrange 4 digits chosen from the set of 7, without repetition, is 840.

Final Answer

The number of integers between 1000 and 9999 that can be formed using the digits {1, 2, 3, 4, 5, 6, 7} without repeating any digit is 840.