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Counting correctly (Posted on 2015-11-19) Difficulty: 3 of 5
Find the number of 7-digit (base-10) numbers, such that each of the digits 1,2,3,4,5 must appear in the number (once or more).

Clearly, no leading zeros.

No Solution Yet Submitted by Ady TZIDON    
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Solution analytic solution, subject to error | Comment 1 of 3
All the numbers must have at least one of each of 1, 2, 3, 4, and 5.

To make 7 digits altogether, two more must be added, possibly duplicating one or two of the five originals.

Possibilities:

00

01, 02, 03, 04, 05

06, 07, 08, 09

11

12, 13, 14, 15

16, 17, 18, 19

22

23, 24, 25

26, 27, 28, 29

33

34, 35

36, 37, 38, 39

44

45

46, 47, 48, 49

55

56, 57, 58, 59

66

67, 68, 69

77

78, 79

88

89

These can be categorized in 8 ways:

Those containing 00:

1234500 can be arranged in 7!/2! ways, but 6! of them will begin with 0.

Start the count at 7!/2! - 6! = 1800


Those containing zero and an extra digit from 1 to 5:

Ex.: 1234501

That can be arranged in 7!/2! ways but 6!/2! begin with 0. But also, there are 5 such sets of digits.

Add to the count (7!/2! - 6!/2!) * 5 = 10800


Those containing zero and an extra digit from 6 to 9:

Ex.: 1234506

There are 7! ways of arranging these, but 6! begin with 0. And there are 4 sets like this.

Add to the count (7! - 6!) * 4 =  17280


Those containing two identical extra digits from 1 to 5:

Ex.: 1234511

There are 7!/3! ways, and there are 5 such sets.

Add to the count 7!/3! * 5 = 4200


Those containing two different extra digits drawn from 1 to 5:

Ex.: 1234512

There are 7!/4 ways and there are C(5,2) = 10 such sets

Add to the count 7!/4 * 10 = 12600



Those containing two identical extra digits from 6 to 9:

Ex.: 1234566

There are 7!/2! ways, and there are 4 such sets.

Add to the count 7!/2! * 4 = 10080


Those containing two different extra digits drawn from 6 to 9:

Ex.: 1234567

There are 7! ways and there are C(4,2) = 6 such sets.

Add to the count 7! * 6 =  30240



Those containing an extra digit from 1 to 5 and one from 4 to 9:

Ex.: 1234516

There are 7!/2! ways and there are 5*4 = 20 such sets.

Add to the count 7!/2! * 20 = 50400



The sum is  137400.

  Posted by Charlie on 2015-11-19 11:13:05
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