Q1: How many combinations can be using 3 alphabet letters [from A to Z]? Note that string doesn't have to be unique, it can be eg. RPP or AZZ
sol: You can make three independent choices, one for each of the three letters. For each choice, you have 26 options [the letters in the alphabet]. So the total number of combinations is
If you want the letters to be unique, the calculation changes slightly. You still have 26 options for the first choice, but for the second choice there are now only 25 options available [all letters except the one you already chose], and for the third choice there are 24 options available [all letters except the two you already chose]. So this gives you:
TASK FOR YOU
Q1: How many possible 7 characters combination using numbers and letters[alphabets]?
You can make three independent choices, one for each of the three letters. For each choice you have 26 options [the letters in the alphabet]. So the total number of combinations is
$$ 26 \cdot 26 \cdot 26 = 26^3 = 17576. $$
If you want the letters to be unique, the calculation changes slightly. You still have 26 options for the first choice, but for the second choice there are now only 25 options available [all letters except the one you already chose], and for the third choice there are 24 options available [all letters except the two you already chose]. So this gives you:
$$ 26 \cdot 25 \cdot 24 = 15600. $$
If letters cannot repeat in a three-letter word, there are $6$ choices [A, B, C, D, E, or F] for the first letter. There are then $5$ choices for the second letter [the five letters that were not chosen in the first letter] and then there are $4$ choices for the third letter. This gives $6 \cdot 5 \cdot 4 = 120$ three-letter words with repeats not allowed.
If letters can be repeated as many times as you want, there are $6$ options [A, B, C, D, E, or F] for the first letter, second letter, and third letter. Then $6^3 = 216$ are the number of options for all three-letter-words.
If a letter can be repeated at most twice, it gets more complicated. Notice that a three-letter word has all different letters, two letters that are the same and one that is different, or all three letters the same. Without any restrictions on the number of repetitions, we found $216$ three-letter words. Of those, there are exactly $6$ which have letters repeated $3$ times [AAA, BBB, CCC, DDD, EEE, and FFF]. That means the other $216 - 6 = 210$ three-letter-words have letters repeated at most twice.
As the name implies, a number system is a mathematical system that is used to represent numerals using various symbols and variables. Under the number system, numbers that can be plotted on a number line, commonly known as real numbers, are represented by a set of values or quantities. Based on their various features, distinct sorts of numbers are classified into different sets or groups. For example, rational numbers are any integers that can be represented in the form p/q, where q is a non-zero integer. Decimal, binary, octal, and hexadecimal are examples of different sorts of systems.
Combinations
It is defined as the process of choosing one, two, or a few elements from a given sequence, regardless of the order in which they appear. If you choose two components from a series that only has two elements to begin with, the order of those elements won’t matter.
Combination Formula
When r items are chosen from n elements in a sequence, the number of combinations is
nCr = n! / r! [n – r]!
For example, let n = 7 and r = 3, then number of ways to select 3 elements out of 7 = 7C3 = 7!/3![7 – 3]! = 35.
How many different 3 letter combinations can be made from Alphabet?
Solution:
If the repetition of letters is allowed, then each alphabet can be chosen from the given 26 alphabets.
Hence total number of combinations = 26C1 × 26C1 × 26C1
= 26 × 26 × 26
= 17576
However, if repetition is not allowed, then the number of combinations = 26C1 × 25C1 × 24C1
= 26 × 25 × 24
= 15600
Similar Problems
Problem 1. Given a piggy bank containing 20 coins, determine the number of permutations of nickels, dimes, and quarters it holds.
Solution:
In this case, the order of the coins clearly does not matter. There’s also no mention of whether or not recurrence is permitted.
Following the formula, we have: n+r−1Cr = n+r−1Cn−1 , in the case of r number of combinations from a sequence with n number of elements where elements can be repeated:
The number of possible pairings = 20+3-1C20 = 22C20
= 22! / [22 – 20]! 20!
= 11[21]
= 231
Problem 2. Tell me how many different methods there are to allocate 7 students to a college trip if we only have one triple room and two double rooms.
Solution:
This problem might be understood as having to divide the seven pupils into three groups of three, two, and two students.
Number of ways to choose 3 students in the triple = 7C3 = 7! / 3!4! = 35
Number of ways to choose 2 out of the remaining 4 students = 4C2 = 4!/ 2!2! = 6
The number of possible ways to choose two students from the remaining two students is equal to one.
Total number of arrangements = 35 × 6 × 1 = 210.
During a conference, 7 students can be assigned to 1 triple and 2 double hotel rooms in 210 different ways.
Problem 3. Determine the number of ways a five-person committee may be established from a group of seven men and six women, with at least three men on the committee.
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 x 6C2] = 525
Number of arrangements when there are 4 men and 1 woman on the committee= [7C4 x 6C1] = 210
Number of arrangements when there are all 5 men on the committee = [7C5] = 21
Total arrangements = 525 + 210 + 21
= 756
Problem 4. Find the number of ways the letters in the word ‘LEADING’ can be arranged so that 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 x 6 = 720.
Problem 5. Find the number of words with four consonants and three vowels that may be made from eight consonants and five vowels.
Solution:
Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = 8C4 x 5C3
= \frac{8 ×7 ×6 ×5 ×4!}{4! × 4!} × \frac{5 ×4 ×3!}{3! × 2!}
= 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.