SOLUTION: How many 7-digit sequences have a digit that appears at least 6 times? (For example, 3339333 and 0200000 are two such sequences. A sequence is allowed to begin with 0.)

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Question 1210205: How many 7-digit sequences have a digit that appears at least 6 times?

(For example, 3339333 and 0200000 are two such sequences. A sequence is allowed to begin with 0.)
Found 3 solutions by CPhill, greenestamps, Edwin McCravy:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let a 7-digit sequence be represented by $d_1 d_2 d_3 d_4 d_5 d_6 d_7$, where each $d_i \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
We are looking for the number of sequences where a digit appears at least 6 times. This can happen in two cases:
Case 1: A digit appears exactly 6 times.
Case 2: A digit appears exactly 7 times.
Case 1: A digit appears exactly 6 times.
First, choose which digit appears 6 times. There are 10 choices (0 to 9).
Next, choose the positions for these 6 digits in the 7-digit sequence. There are $\binom{7}{6}$ ways to choose the 6 positions.
Finally, choose the digit for the remaining position. Since this digit cannot be the same as the digit that appears 6 times, there are 9 choices for this remaining digit.
So, the number of sequences where a digit appears exactly 6 times is $10 \times \binom{7}{6} \times 9 = 10 \times 7 \times 9 = 630$.
Case 2: A digit appears exactly 7 times.
First, choose which digit appears 7 times. There are 10 choices (0 to 9).
Next, all 7 positions in the sequence must be this chosen digit. There is only $\binom{7}{7} = 1$ way to place these digits.
So, the number of sequences where a digit appears exactly 7 times is $10 \times \binom{7}{7} = 10 \times 1 = 10$.
The total number of 7-digit sequences where a digit appears at least 6 times is the sum of the numbers from Case 1 and Case 2.
Total number of sequences = (Number of sequences with exactly 6 repetitions) + (Number of sequences with exactly 7 repetitions)
Total number of sequences = 630 + 10 = 640.
Final Answer: The final answer is $\boxed{640}$
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


At least 6 times means either all seven digits are the same or six of the seven are the same.

(1) all 7 digits the same

The digit can be any one of the 10 digits 0 to 9: 10 choices.

There are 10 7-digit sequences containing a single digit 7 times.

(2) 6 of the 7 digits the same

(a) choose 1 of the 10 digits to be the one that occurs 6 times: 10 choices
(b) choose 1 of the other 9 digits to be the digit that is difference: 9 choices
(c) find the number of arrangements of the string AAAAAAB. That number is



There are 7 ways to arrange those digits.

Total number of 7-digit numbers with 6 of the digits the same: 10*9*7 = 630.

Total number of 7-digit numbers with at least 6 digits the same: 10+630 = 640.

ANSWER: 640
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Case 1: ALL the digits are the same
There are 10 ways all 7 digits can be the same.
That's 10 ways for Case 1.

Case 2:
6 of the digits are the same and 1 is different.

We can choose the digit that occurs 6 times, 10 ways.
We can choose the digit that occur only once, 9 ways.
We can choose the position for the digit that occurs only once, 7 ways.
 
That's (10)(9)(7) = 630 ways for Case 2.

Answer: 10 + 630 = 640 ways.

Edwin
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