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 ways to arrange 6 distinct digits chosen from 7 available digits

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

Answer

50405040

8(ii)

1. Apply the multiplication principle since each of the 6 digit positions can be filled by any of the 7 available digits

Number of ways=7×7×7×7×7×7=76\text{Number of ways} = 7 \times 7 \times 7 \times 7 \times 7 \times 7 = 7^6
76=1176497^6 = 117649

Answer

117649117649

8(iii)

1. Identify the possible frequency combinations of digits that sum to 6, where each digit used appears at least twice.

The 6-digit number must be formed by digits whose frequencies are ≥2\ge 2. We partition 66 into sums of integers ≥2\ge 2:

  • 3 distinct digits appearing 2 times each: 2+2+2=62+2+2=6
  • 2 distinct digits appearing 3 times each: 3+3=63+3=6
  • 2 distinct digits, one appearing 4 times and the other 2 times: 4+2=64+2=6
  • 1 digit appearing 6 times: 66

2. Calculate the number of ways for the 2-2-2 frequency distribution.

For the 2-2-2 distribution:
Number of ways to choose 3 digits from 7: (73)=35\binom{7}{3} = 35
Number of permutations of these 6 digits: 6!2!2!2!=90\frac{6!}{2!2!2!} = 90
Number of numbers formed = 35×90=315035 \times 90 = 3150

3. Calculate the number of ways for the 3-3 and 4-2 frequency distributions.

For the 3-3 distribution:
Number of ways to choose 2 digits from 7: (72)=21\binom{7}{2} = 21
Number of permutations of these 6 digits: 6!3!3!=20\frac{6!}{3!3!} = 20
Number of numbers formed = 21×20=42021 \times 20 = 420

For the 4-2 distribution:
Number of ways to choose which digit appears 4 times and which appears 2 times: 7×6=427 \times 6 = 42
Number of permutations of these 6 digits: 6!4!2!=15\frac{6!}{4!2!} = 15
Number of numbers formed = 42×15=63042 \times 15 = 630

4. Calculate the number of ways for the 6 distribution and find the total sum.

For the 6 distribution:
Number of ways to choose 1 digit from 7: (71)=7\binom{7}{1} = 7
Number of permutations: 6!6!=1\frac{6!}{6!} = 1
Number of numbers formed = 7×1=77 \times 1 = 7

Total possible 6-digit numbers = 3150+420+630+7=42073150 + 420 + 630 + 7 = 4207

Answer

4207

AI analysis — may contain mistakes.

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