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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..

asked in Combinatory by Active (1.9k points) 3 43 80 | 55 views

1 Answer

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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


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