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Nadya [2.5K]
3 years ago
11

A word processing program requires a user to enter a 7-digit registration code made up of the digits 1,2,4,5,6,7 and 9. Each num

ber has to be used, and no number can be used more than once. How many codes are possible?
Mathematics
1 answer:
ahrayia [7]3 years ago
6 0
For the first digit there are 7 choices. For the second digit there are 6 choices (because we can't use the same one again). And so on, until there is only one choice for the last digit.

The number of possible codes is this:
7*6*5*4*3*2*1 = 7! = 5040
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402.12 + 804.24 = 1206.36

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Using only words solve 8x+3=27 step by step​
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Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
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Answer:

1.. Total number of 6 bit strings is 64

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3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

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Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

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6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

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The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

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Answer:

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Step-by-step explanation:

The number 0 is a whole number but not a natural number. But we can say that natural numbers contain all the whole numbers except the number 0.

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