Question

How many bit strings of length 10 have________.a) exactly three 0s?b) more 0s than 1s?c) at least seven 1s?d) at least three 1s slader

Answer 1

Answer: a. 120, b. 386, c. 176, d. 968.

Step-by-step explanation:

For a combination of any number, is given as

C n,r = n!/r!(n-r)!

Please note that "n,r" is a subscript, and the exclamation mark "!" is called factorial.

From the question, n = 10

EXACTLY 3 0s

10 combination 3

r is exactly 3, that is equal 3.

C 10,3= 10!/3!(10-3)! = 10!/3!7!= 120.

For clarification,

10!/3!7!=10×9×8/3×2×1 = 120.

You can also use a calculator to compute the factorials.

MORE 0s than 1s

There will be more 0s than 1s when < 5bits are 0s.

We have r<5

Therefore for r=4

C 10,4 = 10!/4!(10-4)!=10!/4!6!=210

r=3

C 10,3= 10!/3!(10-3)!=10!/3!7!=120

r=2

C 10,2=10!/2!(10-2)!=10!/2!8!=45

r=1

C 10,1=10!/1!(10-1)!=10!/1!9!=10

r=0

C 10,0=10!/0!(10-0)!=10!/0!10!=1

Summing the answers gives us our final answer

210+120+45+10+1= 386.

AT LEAST 7 1s

To get this combination, the value of r will be greater than or equal to 7

r>=7

We have,

r=7

C 10,7=10!/7!(10-7)!=10!/7!3!=120

r=8

C 10,8=10!/8!(10-8)!=10!/8!2!=45

r=9

C 10,9=10!/9!(10-9)!=10!/9!1!=10

r=10

C 10,10=10!/10(10-10)!=10!/10!0!=1

120+45+10+1= 176

AT LEAST 3 1s

the value for r will be greater than or equal to 3:

We can the values of r from 3 to 10.

r=3

C 10,3=10!/3!(10-3)!=120

r=4

C 10,4=10!/4!(10-4)!=10!/4!6!=210

r=5

C 10,5=10!/5!(10-5)!=10!/5!5!=252

r=6

C 10,6=10!/6!(10-6)!=10!/6!4!=210

r=7

C 10,7=10!/7!(10-7)!=10!/7!3!=120

r=8

C 10,8=10!/8!(10-8)!=10!/8!2!=45

r=9

C 10,9=10!/9!(10-9)!=10!/9!1!=10

r=10

C 10,10=10!/10!(10-10)!=10!/10!0!=1

Adding our answers gives 968.

Answer 2

The bits can be either 1 or 0. The total number of bit string for each specified case is:

• Exactly three 0s : 120 strings
• More 0s than 1s: 386 strings
• At least seven 1s: 176 strings
• At least three 1s: 968 strings

How to choose r items out of n indistinguishable items?

Since the items are indistinguishable, their arrangements doesn't matter.

They can be chosen in $^nC_r = \dfrac{n.(n-1).(n-2)...(n-(r+2)).(n-(r+1))}{r.(r-1).(r-2)...3.2.1} \: \rm (r \leq n)$

The bit string is of length 10.

Each bit can be in one of the two states, viz 0 or 1.

Evaluating the count of bit strings for given cases:

• Case 1: Exactly three 0s

Think of it as if there are 10 seats and 3 people to sit on. They're going to be 0s. 3 seats can be chosen from 10 seats in $^{10}C_3 = \dfrac{10\times 9\times 8}{3 \times 2\times 1} = 120$ ways.

The three 0s are identical, so no intra-arrangement between them matters.

Thus, total 120 such strings exist.

• Case 2: More 0s than 1s:

It means, 0s can be 6,7,8,9, or 10 places.

Just similar to above case, 0s on x places out of 10 places can be in $^{10}C_x$ ways.

Thus, total such strings of 0s being more than 1s and being 10 bit strings are:

$^{10}C_6 + ^{10}C_7 + ^{10}C_8 + ^{10}C_9 + ^{10}C_{10} =210+120+45+10+1=386$

• Case 3: At least seven 1s:

At least seven 1s means either 7, 8, 9, or 10 ones.

Total count of such strings are:

$^{10}C_7 + ^{10}C_8 + ^{10}C_9 + ^{10}C_{10} =120+45+10+1=176$

• Case 4: At least three 1s:

They are three or more ones. Total count of such strings is:

$^{10}C_3 + ^{10}C_4 + ^{10}C_5+^{10}C_6 + ^{10}C_7 + ^{10}C_8 + ^{10}C_9 + ^{10}C_{10} =120 + 210 + 252 + 210+120+45+10+1=386+582=968$

Thus, the total number of bit string for each specified case is:

• Exactly three 0s : 120 strings
• More 0s than 1s: 386 strings
• At least seven 1s: 176 strings
• At least three 1s: 968 strings

Learn more about combinations here:

https://brainly.com/question/11958814