How many bit strings of length 10 contains

Question

1. How many bit strings of length 10 contains:
   1. exactly four 1s?
   2. at most four 1s?
   3.  at least four 1s?
   Note: Justify your answers
2. A group contains n men and n women. How many ways are there to arrange these people in a row if the men and women alternate? Justify.
3. How many subsets with more than two elements does a set with 100 elements have? Justify.

Summary

Here we are a total of 3 questions. we have to write their answers one by one. In the first question, we have to answer that how many bit strings of length 10 contain. After that in the second question, we have to find if men and women are in a row then how many ways they can be arranged. In the final third question, we have to find subsets with more than two elements.

Explanation

Permutation and combinations are the processes that are used to arrange the items or the elements in mathematics. But there is a difference between both the processes. So the difference is in Permutation will be used in the case where the order is important. And the combination is used where the order of elements is not important.

In the below formula
n = total number of objects.
r= number of objects selects
Formula for permutation is ,P(n,r)=n!(n-r)!
Formula for combination is, nCr=n!(n-r)!r! n C r = n !(n – r) ! r !

Solution 1:

Exactly 4 ones mean the ones can be at any place. So, there is a total of 210 ways.

At most 4 one’s means any out of 0, 1, 2, 3, 4 number of one’s are possible. so here, if we calculate the ways then
= 1+ 10+ 45+ 120+ 210 = 386 ways

• At least 4 one’s means that a number of one’s in that string can be 10, 9, 8, 7, 6, 5, and 4.
  Here the possibilities of the total ways are as follows
  =1024 – (1+10+45+120)
  =1024- 176
  = 848 ways

Solution 2:

Here we can arrange the n-men and n-women in two different ways.
first, we can arrange the men at odd places and then at the even places so it will look like as:
⦁ Men at odd places, like M W M W M W….
⦁ Men at even places, like W M W M W M….
But the men and women can also interchange their position, so the arrangement process also has plus possibilities.
So it will look like
Possible arrangements= possible arrangement for case (1) + possible arrangements for case (2)
= n! * n! + n! * n!

Solution 3:

Here subset with more than two elements has to be calculated. So,
total subset and subset with at most 2 elements
here is the calculation
so = 1+ 100+ 4950 = 5051
total we have 5051 subsets.