TestBook Test Series: Combinatory - Permutations And Combinations

Question

What is the number of possible bit string of length 7 where number of 1's are more compared to number of 0's??

NOTE: i know any bit string question can be solved using tree diagram..here it will be hell lengthy..so recommend other way.

OR

What is the number of possible bit string of length 7 where number of 1's is atleast four ?

Comments

64??

Answer 1

No. of bit string of length 7 = $2^7 = 128$ as each position can be either 0 or 1.

Lets consider the ones with equal number of 0's and 1's. No such string is possible as string length is odd.

Now, both 1 and 0 are equally likely in a string. This means exactly half of 128 must have more number of 1's and same for more number of 0's. So, our answer must be 64.

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Alternatively, we can say we require either 4 number of 1's or 5 number of 1's, or 6 number of 1's or 7 number of 1's each amounting to ${}^7C_4, {}^7C_5, {}^7C_6$ nd ${}^7C_7$ possibilities. So, 35 + 21 + 7 + 1 = 64.$

Comments

second approach i thought but somehow miscalculated ..thanks for clarification