Binary Numbers

Probability Level 3

A string of binary digits having 0s and 1s is of length 11. How many such strings are possible where the number of 1s in odd places exceeds the number of 1s in even places by 3?


The answer is 165.

Rajen Kapur by Rajen Kapur , 72, India

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1 solution

Rifat Zaman
Mar 9, 2014

No. of odd places: 6

No. of even places: 5

No. of ways

= sum( 6C(3+n) * 5Cn ) [0<=n<=3]

= 165

0 Helpful 0 Interesting 0 Brilliant 0 Confused

i think it should be sum( 6C(4+n) * (sum(5Ck) 0<=k<=n) ) 0<=n<=2

because it is asking for "exceeds"

Dhanpath Sirvee - 7 years, 1 month ago

i mean ans=67

i.e., 6C4 (5C0) + 6C5 (5C0+5C1) + 6C6*(5C0+5C1+5C2)

Dhanpath Sirvee - 7 years, 1 month ago

Can you please check my solution? i have an entirely different answer.

if it is already 11 in length, then it is understood that 1 takes the first odd place. from that there are only 5 places for both odd and even. then, there should only be a difference of two. so,

(5C5 • 5C3)+(5C4•5C2)+(5C3•5C1)+(5C2•5C0)=120

Maymay Uy - 7 years, 1 month ago

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00101010110 is valid

math man - 6 years, 11 months ago

In the solution provided by Rifat Zaman, the summation step gives 20 + 75 + 60 + 10 = 165

Rajen Kapur - 6 years, 7 months ago

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