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Kenneth Rosen Edition 7 Exercise 8.1 Question 25 (Page No. 511)

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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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It will be

$\sum_{_{i = 2k| 0<=k<=3}}\binom{7}{i}$

= $2^{7} / 2$

= $2^{6}$

=$64$
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