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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?
- a)1809
- b)7569
- c)9369
- d)8649
- e)8289
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
How many 6-digit numbers can be formed using the digits {1, 2, 3, ... ...
Key Data
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.
2. Formed using digits {1, 2, 3,..., 9}. Note: Does not include zero.
3. Any digit that appears should appear at least twice.
Examples: Those that satisfy and those that do not
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.,
Some 6-digit numbers that do not satisfy the condition: 123456, 123444, 558812, etc.,
List Down Possibilities and Count
Possibility 1: All 6 digits are same
Example: 111111
9 such numbers possible.
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 6! / 4! × 2! = 15
Number of such numbers = Product of values obtained in the above 3 steps. i.e., 9C2 × 2 × 15
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 6! / 4! × 2! = 15
Number of such numbers = Product of values obtained in the above 3 steps. i.e., 9C2 × 2 × 15
= 
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
Example: 444777
Step 1: We are selecting 2 digits from 9 numbers. This can be done in 9C2
=
ways
Step 2: Reordering of 6 digits can be done in 6!3!×3!6!3!×3! = 6×5×4×3!3!×3!6×5×4×3!3!×3! = 20 ways
Number of such numbers = 36 × 20 = 720
waysStep 2: Reordering of 6 digits can be done in 6!3!×3!6!3!×3! = 6×5×4×3!3!×3!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
Step 1: We are selecting 3 digits from 9 numbers. This can be done in 9C3
= 
= 84 ways.
Step 2: Reordering of 6 digits can be done in
= 90 ways
Number of such numbers = 84 × 90 = 7560
= 84 ways.Step 2: Reordering of 6 digits can be done in
= 90 waysNumber of such numbers = 84 × 90 = 7560
Total such numbers = (9 + 1080 + 720 + 7560) = 9369
Choice C is the correct answer.
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Community Answer
How many 6-digit numbers can be formed using the digits {1, 2, 3, ... ...
To solve this problem, we can break it down into smaller cases and then add up the total number of possibilities for each case. Let's consider the different cases:
Case 1: All digits are distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, 7 choices for the third digit, and so on. Therefore, the total number of possibilities is 9 * 8 * 7 * 6 * 5 * 4 = 12,960.
Case 2: Two digits are repeated twice, and the other three digits are distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth and fifth digits must be one of the two digits that are repeated twice, which can be chosen in 9 ways. The sixth digit must be the remaining digit that has not been used yet, which can be chosen in 8 ways. Therefore, the total number of possibilities is 9 * 8 * 7 * 9 * 9 * 8 = 36,288.
Case 3: Three digits are repeated twice, and the other two digits are distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth and fifth digits must be two of the three digits that are repeated twice, which can be chosen in 9 ways. Therefore, the total number of possibilities is 9 * 8 * 7 * 9 * 8 = 36,288.
Case 4: Four digits are repeated twice, and the other digit is distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth and fifth digits must be two of the four digits that are repeated twice, which can be chosen in 9 ways. Therefore, the total number of possibilities is 9 * 8 * 7 * 9 = 4,536.
Case 5: Five digits are repeated twice
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth, fifth, and sixth digits must be three of the five digits that are repeated twice, which can be chosen in 9 ways. Therefore, the total number of possibilities is 9 * 8 * 7 = 504.
Adding up the possibilities from all the cases, we get 12,960 + 36,288 + 36,288 + 4,536 + 504 = 90,576.
However, we need to subtract the cases where all the digits are the same (e.g., 111111), which is only 9 possibilities.
Therefore, the total number of 6-digit numbers that can be formed is 90,576 - 9 = 90,567.
Hence, the correct answer is option 'C' (9369).
Case 1: All digits are distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, 7 choices for the third digit, and so on. Therefore, the total number of possibilities is 9 * 8 * 7 * 6 * 5 * 4 = 12,960.
Case 2: Two digits are repeated twice, and the other three digits are distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth and fifth digits must be one of the two digits that are repeated twice, which can be chosen in 9 ways. The sixth digit must be the remaining digit that has not been used yet, which can be chosen in 8 ways. Therefore, the total number of possibilities is 9 * 8 * 7 * 9 * 9 * 8 = 36,288.
Case 3: Three digits are repeated twice, and the other two digits are distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth and fifth digits must be two of the three digits that are repeated twice, which can be chosen in 9 ways. Therefore, the total number of possibilities is 9 * 8 * 7 * 9 * 8 = 36,288.
Case 4: Four digits are repeated twice, and the other digit is distinct
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth and fifth digits must be two of the four digits that are repeated twice, which can be chosen in 9 ways. Therefore, the total number of possibilities is 9 * 8 * 7 * 9 = 4,536.
Case 5: Five digits are repeated twice
In this case, we have 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. The fourth, fifth, and sixth digits must be three of the five digits that are repeated twice, which can be chosen in 9 ways. Therefore, the total number of possibilities is 9 * 8 * 7 = 504.
Adding up the possibilities from all the cases, we get 12,960 + 36,288 + 36,288 + 4,536 + 504 = 90,576.
However, we need to subtract the cases where all the digits are the same (e.g., 111111), which is only 9 possibilities.
Therefore, the total number of 6-digit numbers that can be formed is 90,576 - 9 = 90,567.
Hence, the correct answer is option 'C' (9369).
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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?a)1809b)7569c)9369d)8649e)8289Correct answer is option 'C'. Can you explain this answer?
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