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 many 6-bit strings are there? How many 6-bit strings have weight 0? How many 6-bit strings have weight 1? How many 6-bit strings have weight 3? How many 6-bit strings have weight 5? How many 6-bit strings have weight 6? How many 6-bit strings have weight 7?

Question

Madyson Santana

Mathematics

28 September, 11:04

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 many 6-bit strings are there? How many 6-bit strings have weight 0? How many 6-bit strings have weight 1? How many 6-bit strings have weight 3? How many 6-bit strings have weight 5? How many 6-bit strings have weight 6? How many 6-bit strings have weight 7?

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Answers (1)

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Augustus Rich · Answer 1

1 ... Total number of 6 bit strings is 64

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

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!

6C0 = 1

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

Hence, the number of string with weight 5 is 6

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

Weight of 7 means that a string that has 7 occurrence of 1

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.