Question

Using the letters of the word "ENGLISH", how many five letters words can begin with G? (1) 2520 (2) 360 (3) 180 (4) 1260

Short answer

Answer: 360

Step-by-step solution

Identify the problem type

We are trying to create a five-letter word using the letters of the word "ENGLISH" starting with 'G'. This is a problem of permutation since the order in which the remaining letters are placed matters.

Calculate the possible selections

There are 7 unique letters in the word "ENGLISH". As the word should begin with 'G', we are left with 6 letters to select from to fill the remaining 4 positions.

Apply the permutation formula

We have 6 letters and 4 positions to fill, so we will apply the permutation formula: P(n, r) = n! / (n - r)! Where n is the total number of unique items to select from, r is the number of positions available, and ! denotes the factorial. In this case, n = 6 and r = 4. P(6, 4) = 6! / (6-4)!

Calculate the factorials

For solving the permutation formula, we will calculate the factorials value of 6 and (6-4). 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 (6 - 4)! = 2! = 2 × 1 = 2

Solve the permutation formula

Now, we will substitute the factorial values in the permutation formula: P(6, 4) = 720 / 2 = 360

Conclusion

There are 360 possible five-letter words that can begin with 'G' using the letters of the word "ENGLISH". Therefore, the correct answer is option (2) 360.

Key concepts

Factorial Notation

In mathematics, factorial notation is an essential concept, particularly in combinatorics and probability theory. It is symbolized by an exclamation point (!) and represents the product of an integer with all the positive integers below it. For example, the factorial of 4, denoted as 4!, is calculated as:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
Factorial zero, represented as 0!, is defined to have a value of 1. One important property of factorial values is that they grow exceptionally fast, which is critically relevant when calculating the number of possible permutations or combinations in a given set. The concept of factorials is profoundly linked to permutations since factorials provide the total count of possible arrangements of a set of objects when order matters.

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is closely related to other fields such as algebra, geometry, and probability theory. Combinatorics encompasses several subfields, including graph theory, enumeration, and design theory, each concerned with different types of problem-solving approaches and applications.
One of the fundamental principles in combinatorics is the counting principle, which enables the calculation of the number of possible outcomes of various scenarios. For example, in our solved exercise, combinatorics helps in determining the number of different words that can be formed with a given set of letters. Combinatorics provides tools like permutations and combinations to efficiently solve problems without having to list all possible outcomes.

Arrangements and Permutations

Arrangements and permutations are at the heart of many combinatorial problems. The terms often get used interchangeably, but they serve the purpose of counting and arranging objects where the order is significant.
In the context of permutations, we use a specific formula when we want to arrange a subset of a larger set. As shown in our exercise example, to find how many five-letter words you can create from the letters of 'ENGLISH' starting with 'G,' you use the permutation formula:
\[ P(n, r) = \frac{n!}{(n-r)!} \]
where \( n \) is the total number of items to choose from, and \( r \) is the number of items to arrange. In the case of the exercise, we had to arrange 4 out of 6 remaining letters after fixing 'G' as the first letter, which led us to calculate \( P(6, 4) \). The detailed solution steps broke down how this formula led to the answer, emphasizing how understanding permutations is crucial to solving such problems in combinatorics.