How many different 3 letter combinations can be made from Alphabet?

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

ⁿC_r = 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−1C_r = ^n+r−1C_n−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.