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

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.

15 units use socratic

step-by-step explanation: 15 units