To determine the number of six-digit numbers that can be formed from the given sets of digits, we use the formula for permutations of multiset elements:

$\text{Total permutations} = \frac{n!}{k_1! k_2! \dots k_m!}$

where:

• $n$ is the total number of digits,
• $k_1, k_2, \dots, k_m$ are the frequencies of the distinct digits.

(a) Digits: 1,1,1,1,1,3

Here, there are 6 digits in total: five 1's and one 3.

$\text{Total permutations} = \frac{6!}{5!1!} = \frac{720}{120} = 6$

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(b) Digits: 2,2,2,7,7,7

Here, there are 6 digits in total: three 2's and three 7's.

$\text{Total permutations} = \frac{6!}{3!3!} = \frac{720}{6 \times 6} = \frac{720}{36} = 20$

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(c) Digits: 5,6,6,6,7,7

Here, there are 6 digits in total: one 5, three 6's, and two 7's.

$\text{Total permutations} = \frac{6!}{1!3!2!} = \frac{720}{1 \times 6 \times 2} = \frac{720}{12} = 60$

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(d) Digits: 8,8,9,9,9,9

Here, there are 6 digits in total: two 8's and four 9's.

$\text{Total permutations} = \frac{6!}{2!4!} = \frac{720}{2 \times 24} = \frac{720}{48} = 15$

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Final Answers:

(a) 6
(b) 20
(c) 60
(d) 15

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Would you like a step-by-step breakdown of the calculations?

Related Questions:

1. How does the formula for permutations of multiset elements work?
2. What is the difference between permutations and combinations?
3. How would the answers change if the numbers could be repeated?
4. What if the order of digits did not matter?
5. How can factorial calculations be simplified in such problems?

Tip:

When solving problems involving permutations of repeated elements, always simplify factorials before multiplying to avoid large numbers.