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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

$T_{1}$ (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..

+1 vote

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 and at most two 0s

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

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

= 1024 - 112

= 912

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
0
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