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: 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 = 10EXACTLY 3 0s10 combination 3r 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 1sThere will be more 0s than 1s when < 5bits are 0s.We have r<5Therefore for r=4C 10,4 = 10!/4!(10-4)!=10!/4!6!=210r=3C 10,3= 10!/3!(10-3)!=10!/3!7!=120r=2C 10,2=10!/2!(10-2)!=10!/2!8!=45r=1C 10,1=10!/1!(10-1)!=10!/1!9!=10r=0C 10,0=10!/0!(10-0)!=10!/0!10!=1Summing the answers gives us our final answer210+120+45+10+1= 386.AT LEAST 7 1sTo get this combination, the value of r will be greater than or equal to 7r>=7We have,r=7C 10,7=10!/7!(10-7)!=10!/7!3!=120r=8C 10,8=10!/8!(10-8)!=10!/8!2!=45r=9C 10,9=10!/9!(10-9)!=10!/9!1!=10r=10C 10,10=10!/10(10-10)!=10!/10!0!=1120+45+10+1= 176AT LEAST 3 1sthe value for r will be greater than or equal to 3:We can the values of r from 3 to 10.r=3C 10,3=10!/3!(10-3)!=120r=4C 10,4=10!/4!(10-4)!=10!/4!6!=210r=5C 10,5=10!/5!(10-5)!=10!/5!5!=252r=6C 10,6=10!/6!(10-6)!=10!/6!4!=210r=7C 10,7=10!/7!(10-7)!=10!/7!3!=120r=8C 10,8=10!/8!(10-8)!=10!/8!2!=45r=9C 10,9=10!/9!(10-9)!=10!/9!1!=10r=10C 10,10=10!/10!(10-10)!=10!/10!0!=1Adding our answers gives 968.