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?
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 9C2 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., 9C2 × 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 9C2 = \\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 9C3 = \\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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