Question

The first 12 letters of the english alphabet are to be written at random so that there are exactly 4 letters between \(\mathrm{A}\) and \(\mathrm{B}\). the number of ways this can be done is (a) \(7 \cdot 10 !\) (b) \(2 \cdot 10 !\) (c) \(21 \cdot 10 !\) (d) \(14 \cdot 10 !\)

Short answer

The number of ways the first 12 letters of the English alphabet can be written at random so that there are exactly 4 letters between A and B is \(\boxed{(d) \ 14 \cdot 10!}\).

Step-by-step solution

Determine the number of slots available for A and B

We can either have A first and then B, or B first and then A. In either case, since there must be exactly 4 letters between A and B, we will find the overall slots available for A and B. We have 12 letters, thus 12 positions. If A is placed first, 4 letters will follow, and then B must be placed. Thus, there are 7 possible positions for A (since we need to leave room for the next 4 letters and B in the subsequent positions). However, if B comes before A, the same logic applies, so we have 7 positions for B. So, in total, there are 2 * 7 = 14 ways to place A and B.

Arrange the remaining 10 letters

Once we have placed A and B in their appropriate positions (in one of the 14 ways we found in Step 1), we have 10 remaining letters to arrange in the remaining positions. To arrange the 10 remaining letters, there are \(10!\) possible ways.

Combine the number of ways for A and B with the remaining letters

In Step 1, we found there are 14 possible ways to place A and B. And in Step 2, we found there are \(10!\) ways to arrange the remaining 10 letters. To get the total number of ways to arrange all first 12 alphabet letters with the given constraints, we multiply these two values together: \(14 \cdot 10!\) Therefore, the answer is \(\boxed{(d) \ 14 \cdot 10!}\).

Key concepts

Combinatorics

Combinatorics is a branch of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties related to counting. It is related to all sorts of activities that involve arrangement or selection of objects, where the order can be significant or not.

For instance, say you have a set of distinct books and you want to find out in how many different ways you can arrange them on a shelf. This is a problem of permutations, a fundamental concept within combinatorics, where the order matters. Combinatorics also covers combinations, where you select items from a group without regard to the order in which they're selected.

Understanding combinatorics is invaluable when solving problems like the given textbook exercise, where we have to determine the number of ways we can arrange letters under certain conditions. This kind of problem also illustrates the practical importance of combinatorics in mathematical and real-world problems, such as coding theory and the organization of complex systems.

Factorial Notation

Factorial notation is a mathematical expression used frequently in combinatorics to simplify the description of permutations and combinations. The factorial of a positive integer, denoted by an exclamation point (!), is the product of all positive integers less than or equal to that number. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

Zero factorial is defined to be 1, symbolized as \(0! = 1\). It might seem odd at first, but it's useful in allowing the definitions of combinations and permutations to be consistent, even when selecting no items.

Factorials grow extremely fast with increasing numbers, so they are instrumental in quantifying the huge numbers of ways objects can be arranged. As shown in the exercise's solution, \(10!\) – the factorial of 10 – represents the number of ways to arrange 10 different letters, which in itself is an astronomical number (over 3 million possibilities). Factorial notation is thus a concise and effective way to represent these large numbers.

Arrangement of Objects

The arrangement of objects, often referred to as permutations, is a key topic in combinatorics that allows us to tally the number of distinct ways to organize a set of items. When you permute items, you're counting the different ways they can be ordered.

For example, consider arranging letters in a word. If the word is 'CAT,' there are 6 different permutations: CAT, CTA, ATC, ACT, TCA, and TAC. We calculate this by considering that there are 3 choices for the first letter, 2 for the second, and only 1 for the last, which leads us to \(3! = 6\) arrangements. This calculation can be scaled up for any number of objects.

Returning to our textbook exercise, once A and B were allotted specific places, the arrangement of the remaining letters becomes a simpler permutation problem, where the remaining 10 letters can be arranged in any order, totaling to \(10!\) different permutations. This demonstrates how permutation problems vary in complexity depending on imposed conditions, like separating certain items by a fixed number of spaces.