Question

Binary is a sequence of 0s and 1s such as 1001101 which can be used to code information. To break a code, you need to find how many 0s and 1s there are in a particular sequence of 7 numbers. A break in the code shows the permutation of the symbols is 35. How many many 0s and how many 1s are there if you know there are more 1s than 0s

Answer 1

Answer:

There are four 1s and three 0s

Explanation:

First, we know that there are 7 numbers in the sequence and there are more 1s than 0s, so we have the following options:

Option 1. Seven 1s and zero 0s

Option 2. Six 1s and one 0s

Option 3. Five 1s and two 0s

Option 4. Four 1s and three 0s

On the other hand, the number of permutations for n elements that are not all different is calculated as:

$\frac{n!}{n_1!*n_2!*...+n_k!}$

Where k is the number of elements that are different and $n_i$ is the number of times that every element appears.

Now, we know that the permutation of the symbols is 35, so we need to calculated the number of permutations for every option.

Then, for option 1, we have 7 elements (7 numbers of the sequence) with 2 different elements (seven 1s and zero 0s). So, the number of permutation of the symbols is:

Option 1. $\frac{7!}{7!*0!}=1$

Because n is equal to 7, k is equal to 2 , $n_1$ is equal to 7 and $n_2$ is equal to 0. At the same way we have:

For option 2. $\frac{7!}{6!*1!}=7$

For option 3. $\frac{7!}{5!*2!}=21$

For option 4. $\frac{7!}{4!*3!}=35$

It means that the option in which the permutation of the symbols is 35 is Option 4, so there are four 1s and three 0s.