CASE 1: number of 0s is 0 (all 1s) $= {}^7C_1 = 1$ CASE 2: number of 0s is 2 $= {}^7C_2 = 21$ CASE 3: number of 0s is 4 $= {}^7C_4 = 35$ CASE 4: number of 0s is 6 $= {}^7C_6 = 7$ So, total number of bit sequences possible $= 1+ 21+ 35 + 7 = 64.$
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