The given question is a challenging GMAT 700 800 level quant problem solving question combining concepts in permutation combination (counting methods) and number properties. Another interesting practice question to understand how to list down possibilities and then count outcomes for each of those possibilities.

Question 23: How many 6-digit numbers can be formed using the digits {1, 2, 3, ... 9} such that any digit that appears in such a number appears at least twice?

- 1809
- 7569
- 9369
- 8649
- 8289

@ INR

1. 6-digit numbers.

2. Formed using digits {1, 2, 3,..., 9}. __Note__: Does not include zero.

3. Any digit that appears should appear at least twice.

Some 6-digit numbers that satisfy the condition: 555555, 223344, 111999, etc.,

Some 6-digit numbers that do not satisfy the condition: 123456, 123444, 558812, etc.,

**Possibility 1**: All 6 digits are same

**Example**: 111111

**9** such numbers possible.

**Possibility 2**: 4 digits show one value and 2 digits show another value. Example: 373777

**Step 1**: We are selecting 2 digits from 9 numbers. This can be done in 9C_{2} ways.

**Step 2**: For example, if the digits are 3 and 7, either 3 appears 4 times and 7 appears twice or vice versa.

So, there are 2 possibilities.

**Step 3**: Reordering of 6 digits can be done in \\frac{6!}{4! × 2!}) = 15

Number of such numbers = Product of values obtained in the above 3 steps. i.e., 9C_{2} × 2 × 15

= \\frac{9 × 8}{1 × 2}) × 2 × 15 = **1080**

**Possibility 3**: 3 digits show one value and another 3 digits show a second value.

**Example**: 444777

**Step 1**: We are selecting 2 digits from 9 numbers. This can be done in 9C_{2} = \\frac{9 × 8}{1 × 2}) = 36 ways.

**Step 2**: Reordering of 6 digits can be done in \\frac{6!}{3! × 3!}) = \\frac{6 × 5 × 4 × 3!}{3! × 3!}) = 20 ways

Number of such numbers = 36 × 20 = **720**

**Possibility 4**: 3 different digits, each appearing twice. Example: 234234

**Step 1**: We are selecting 3 digits from 9 numbers. This can be done in 9C_{3} = \\frac{9 × 8 × 7}{1 × 2 × 3}) = 84 ways.

**Step 2**: Reordering of 6 digits can be done in \\frac{6!}{2! × 2! × 2!}) = \\frac{720}{8}) = 90 ways

Number of such numbers = 84 × 90 = **7560**

Total such numbers = (9 + 1080 + 720 + 7560) = **9369**

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