How many six-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 if

  1. ((i))

    repetition of digits is not allowed?

  2. ((ii))

    repetition of digits is allowed?

  3. ((iii))

    every digit in the number appears at least twice?

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Worked solution

8((i))

1. Calculate the number of permutations for choosing and arranging 6 digits from 7 distinct digits

Number of ways = 7P6^7P_6

7P6=7×6×5×4×3×2=5040^7P_6 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 5040

Answer

50405040

8((ii))

1. Determine the number of choices for each digit and apply the multiplication principle

There are 77 available digits. Since repetition is allowed, each of the 66 positions in the number can be filled in 77 ways.
Number of ways=76\text{Number of ways} = 7^6

2. Evaluate the expression

76=1176497^6 = 117649

Answer

117649117649

8((iii))

1. Identify the possible frequency distributions for the digits making up the 6-digit number.

Since the number has 6 digits and every digit present must appear at least twice, the possible frequency distributions (partitions of 6 into parts ≥2\ge 2) are:

  • Case 1: One digit appears 6 times (freq: 66)
  • Case 2: One digit appears 4 times, another appears 2 times (freq: 4,24, 2)
  • Case 3: Two distinct digits appear 3 times each (freq: 3,33, 3)
  • Case 4: Three distinct digits appear 2 times each (freq: 2,2,22, 2, 2)

2. Calculate the number of valid arrangements for Cases 1 and 2.

Case 1 (6):
Number of ways to choose the digit = (71)=7\binom{7}{1} = 7
Arrangements of this digit = 6!6!=1\frac{6!}{6!} = 1
Number of ways = 7×1=77 \times 1 = 7

Case 2 (4, 2):
Choose 1 digit to appear 4 times and 1 digit to appear 2 times: (71)×(61)=42\binom{7}{1} \times \binom{6}{1} = 42 ways.
Arrangements = 6!4!2!=15\frac{6!}{4!2!} = 15
Number of ways = 42×15=63042 \times 15 = 630

3. Calculate the number of valid arrangements for Cases 3 and 4.

Case 3 (3, 3):
Choose 2 digits: (72)=21\binom{7}{2} = 21 ways (not 7×67 \times 6, as the frequencies are identical, order of choosing does not matter).
Arrangements = 6!3!3!=20\frac{6!}{3!3!} = 20
Number of ways = 21×20=42021 \times 20 = 420

Case 4 (2, 2, 2):
Choose 3 digits: (73)=35\binom{7}{3} = 35 ways.
Arrangements = 6!2!2!2!=90\frac{6!}{2!2!2!} = 90
Number of ways = 35×90=315035 \times 90 = 3150

4. Sum the possibilities to find the total number of 6-digit numbers.

Total valid numbers=7+630+420+3150=4207\text{Total valid numbers} = 7 + 630 + 420 + 3150 = 4207

Answer

42074207

AI analysis — may contain mistakes.

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