N Choose K Formula: Combination is described as the process of choosing one, two, or a few elements from a given sequence, independent of the order in which they appear. If you pick two components from a series that only contains two elements, to begin with, the order of those elements won’t matter.
In this article, we will learn about N Choose K Formula also known as Combinations Formula.
N Choose K Formula
When r items are chosen from n elements in a sequence, the number of combinations is
nCk =
For example, let n = 5 and r = 2, then the number of ways to choose two components from a set of five = 5C2 = 5! / 2! (5 – 2)! = 10.
It’s worth mentioning that if r number of combinations are to be obtained from a collection of n components, and such elements can be repeated, then
n+r−1Cr = n+r−1Cn−1
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Similar Problems on N Choose K Formula
Question 1. How many binary strings of length 5 have exactly two 1’s somewhere in the string?
Solution:
Note that the order of the bit is not important in this case because we are concerned with the number of ones in the said string and not their order. Thus, we need to apply the concept of combinations to find the required value.
Here, n = 5 and r = 2.
C(5, 2) =
= 10
So there are 10 bit strings of length 5 with exactly two 1’s in them.
Question 2. Find the number of ways in which a committee of five persons can be formed if they are to be selected from a group of 7 men and 6 women, so as to have at least 3 men on there.
Solution:
At least three men on the committee means we can have either exactly three, four or all five men in the committee.
Number of arrangements when there are 3 men and 2 women on the committee = (7C3 × 6C2) = 525
Number of arrangements when there are 4 men and 1 woman on the committee= (7C4 × 6C1) = 210
Number of arrangements when there are all 5 men on the committee = (7C5) = 21
Total arrangements = 525 + 210 + 21
= 756
Question 3. Find the number of arrangements of the letters of the word ‘LEADING’ where the vowels always appear together?
Solution:
If the vowels are to appear together, they would form a separate letter in the word. Hence we are left with 4 + 1 = 5 letters, which can be arranged in 5! = 120 ways.
Furthermore, there are 3! = 6 ways to arrange the vowels together.
Total number of ways of arranging the letters = 120 × 6 = 720.
Question 4. Find the number of words having 4 consonants and 3 vowels which can be formed out of 8 consonants and 5 vowels.
Solution:
Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = 8C4 × 5C3
=
= 70 × 10 = 700
Number of ways of arranging the 7 letters among themselves = 7! = 5040
Number of words that can be formed = 5040 × 700 = 3528000.
Question 5. How many bit strings of length 8 have an equal number of O’s and 1’S?
Solution:
You are choosing from a set of eight symbols {1, 1, 1, 1, 0, 0, 0, 0} (which would normally give 8! = 40320 choices, but you have three identical “1″s and three identical “0”s so that reduces the number of options to
=
= 70 bit- strings
Question 6. How many strings of 8-bit have at least two consecutive O’s or two consecutive 1’s?
Solution:
An 8 bit string could represent all numbers between 0 and 28 = 256.
Two consecutive zeroes can start at position 1, 2, 3, 4, 5, 6 or 7.
Starting at position 1: Strings of the form [0 0 x x x x x x]
Remaining 6 places can be arranged in 26 = 64 ways.
Starting at position 2: Strings of the form [1 0 0 x x x x x]
Remaining 5 places can be arranged in 25 = 32 ways.
Similarly, the count for positions 3, 4, 5, 6 and 7 comes to be 32 each.
Number of ways = 64 + 32 + 32 + 32 + 32 + 32 + 32 + 32 – 10
= 246
Also number of strings that start with 2 consecutive 1’s = 246
Thus required value = 246 + 246 = 492
Question 7. How many eight-bit strings contain at least two 0s in a row?
Solution:
The total number of 8−bit strings (Since, you are talking about ‘strings’, I assume leading zeroes are valid) = 256
The total number of strings where we don’t have consecutive 0’s = 9C0 + 8C1 + 7C2 + 6C3 + 5C4
= 55
The number of 8 bit strings where we have consecutive 0 s = 256 – 55
= 201