(i) To find the number of different 6-digit numbers, we use the formula for permutations of multiset:
\(\frac{6!}{3!}\)
where 6 is the total number of digits and 3 is the number of repeated digits (the digit 7 appears three times).
Calculating gives:
\(\frac{6!}{3!} = \frac{720}{6} = 120\)
Thus, there are 120 different 6-digit numbers.
(ii) We need to count the numbers that start and end with an odd digit. The odd digits available are 5 and 7.
Case 1: Start with 5 and end with 7.
We arrange the remaining digits 4, 6, 7, 7. The number of arrangements is:
\(\frac{4!}{2!} = \frac{24}{2} = 12\)
Case 2: Start with 7 and end with 5.
We arrange the remaining digits 4, 6, 7, 7. The number of arrangements is:
\(\frac{4!}{2!} = \frac{24}{2} = 12\)
Case 3: Start with 7 and end with 7.
We arrange the remaining digits 4, 5, 6. The number of arrangements is:
\(4! = 24\)
Adding all cases gives:
\(12 + 12 + 24 = 48\)
Thus, there are 48 numbers that start and end with an odd digit.
