Expert Answer to "How many bit strings of length 10 have..."

High School
Algebra
Pre-Algebra
Factors and multiples

Wotzdorfg

Answered question

2021-10-27

How many bit strings of length 10 have at least three 1s?

Answer & Explanation

Alara Mccarthy

Skilled2021-10-28Added 85 answers

n=10
$r \geq 3$
A combination's definition should be evaluated:
$C (10, 3) = \frac{10!}{3! (10 - 3)!} = \frac{10!}{3! 7!} = 120$
$C (10, 4) = \frac{10!}{4! (10 - 4)!} = \frac{10!}{4! 6!} = 210$
$C (10, 5) = \frac{10!}{5! (10 - 5)!} = \frac{10!}{5! 5!} = 252$
$C (10, 6) = \frac{10!}{6! (10 - 6)!} = \frac{10!}{6! 4!} = 210$
$C (10, 7) = \frac{10!}{7! (10 - 7)!} = \frac{10!}{7! 3!} = 120$
$C (10, 8) = \frac{10!}{8! (10 - 8)!} = \frac{10!}{8! 2!} = 45$
$C (10, 9) = \frac{10!}{9! (10 - 9)!} = \frac{10!}{9! 1!} = 10$
$C (10, 10) = \frac{10!}{10! (10 - 10)!} = \frac{10!}{10! 0!} = 1$
Add the number of bit strings for each value of r:
120+210+252+210+120+45+10+1=968

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