edited by
1,821 views
4 votes
4 votes

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

My Approach:->

using product rule

 There are 3 subtask following

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..

edited by

3 Answers

Best answer
2 votes
2 votes

Bit strings of length that 10 contain at least three 1s and at least three 0s  = Total strings possible with 10 bits - Bit strings of length 10 that contains at most two 1s or at most two 0s

= 210 - 2 * (10 choose 2 + 10 choose 1 + 10 choose 0)

= 1024 - 2 * (45 + 10 + 1)

= 1024 - 112

= 912

edited by
2 votes
2 votes
number of 1s>=3 AND number of 0s>=3

complement of (number of 1s>=3 AND number of 0s>=3)

=number of 1s<3 or number of 0s<3

=number of 1s={0,1,2} or number of 0s={0,1,2}

=10C0+10C1+10C2+10C0+10C1+10C2

=112

total number of strings with length 10=$2^{10}$

answer=$2^{10}-112$

=1024-112=912
0 votes
0 votes
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

Related questions

4 votes
4 votes
4 answers
3
2 votes
2 votes
1 answer
4
aditi19 asked Nov 16, 2018
1,913 views
How many solutions are there to the equation x1+x2+x3=17 with x1<6, x3>5?