+1 vote
55 views

How many bit strings of length 10 contain at least three 1s and at least three 0s?

My Approach:->

using product rule

$T_{1}$ (filling 3 ones in 10 places) =$\binom{10}{3}$

$T_{2}$ (filling 3 zeros in remaing 7 places) =$\binom{7}{3}$

$T_{3}$ (filling remaining 4 places) = $2^{4}$

So,total number of bit string = $\binom{10}{3}$*$\binom{7}{3}$*$2^{4}$   which is greater than  $2^{10}$(total number of string).

Now , i want to know what is wrong in my apporach. please explain..

if we consider the complementary problem: at most 2 0s or at most 2 1s. The counts for each case is (10c2)+(10c1)+(10c0)=56 the count for at least 3 0s and at least 3 1s is therefore 2^(10)−2⋅56=912
answered by Loyal (3.2k points) 2 11 28
you are considering one approach here only 3 ones  and 3 zeroes case here question is about atleast  try to approach these problems with complementary method always here

+1 vote
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}