Question

In how many ways can 3 consonants be selected from the English alphabet? (1) \({ }^{21} \mathrm{C}_{3}\) (2) \({ }^{26} \mathrm{C}_{3}\) (3) \({ }^{21} \mathrm{C}_{\mathrm{s}}\) (4) \({ }^{266} \mathrm{C}_{5}\)

Short answer

Answer: There are 1330 ways to select 3 consonants from the English alphabet.

Step-by-step solution

Identify the number of consonants in the English alphabet

There are 26 letters in the English alphabet, and 5 of them are vowels. Therefore, 26 - 5 = 21 consonants.

Use the combination formula to find the number of ways to select 3 consonants

We will use the combination formula, which is written as: \(\ ^n\mathrm{C}_r = \frac{n!}{r!(n-r)!}\) In this case, n = 21 (the number of consonants) and r = 3 (the number of consonants to be selected).

Substitute the values into the combination formula and solve

Plug the values of n and r into the combination formula: \(\ ^{21}\mathrm{C}_3 = \frac{21!}{3!(21-3)!}\) Simplify the expression: \(\ ^{21}\mathrm{C}_3 = \frac{21!}{3! \cdot 18!}\) Now, calculate the factorials: \(21! = 21 \cdot 20 \cdot 19 \cdot 18!\) \(3! = 3 \cdot 2 \cdot 1 = 6\) Substitute the factorials back into the expression: \(\ ^{21}\mathrm{C}_3 = \frac{21 \cdot 20 \cdot 19 \cdot 18!}{6 \cdot 18!}\) The 18! terms cancel out: \(\ ^{21}\mathrm{C}_3 = \frac{21 \cdot 20 \cdot 19}{6}\) Now, do the arithmetic: \(\ ^{21}\mathrm{C}_3 = \frac{7980}{6} = 1330\) The number of ways to select 3 consonants from the English alphabet is 1330. The correct answer is (1) \({ }^{21} \mathrm{C}_{3}\).

Key concepts

Permutations and Combinations

Permutations and combinations are fundamental concepts in combinatorics, a branch of mathematics concerned with counting, arrangement, and possibility in a set. What makes them distinct is how they deal with arrangement and selection.

Permutations focus on the arrangement of a set of items where the order matters. For instance, how many different ways can you arrange the letters A, B, and C? The answer is 6, which includes ABC, ACB, BAC, BCA, CAB, and CBA.

Combinations, on the other hand, are about selection where the order doesn't matter. For example, if you are selecting 2 fruits from an apple (A), banana (B), and cherry (C), the possibilities are AB, BC, and AC. Whether you choose AB or BA doesn't matter; it's the same combination of two fruits.

In the exercise question, selecting 3 consonants from the English alphabet does not concern the order in which they are selected. Hence, we use combinations to solve it. Understanding the difference between these two concepts is vital for solving problems related to counting and probability.

Factorial Notation

Factorial notation is a mathematical concept used to simplify the counting of permutations and combinations. It is denoted by an exclamation point (!) and represents the product of all positive integers up to a given number.

For instance, the factorial of 5, written as \(5!\), is \(5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials are integral to calculating combinations because they allow us to easily determine the total number of ways to arrange or select items.

In the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), we see the factorial notation at work. Simplifying factorials carefully when some terms cancel out, as seen in the given solution, is essential for accurate calculation. For example, you'll notice that in \( \frac{21!}{3! \cdot 18!} \), the term \(18!\) appears in both numerator and denominator, and therefore cancels out, significantly simplifying our calculations.

Probability

Probability is the measure of how likely an event is to occur and is often expressed as a fraction or percentage. It ranges from 0 (the event will not occur) to 1 (the event will certainly occur).

To calculate the probability of an event, you can use the ratio of the number of favorable outcomes to the total number of possible outcomes. In a combination problem, if you were asked to find the probability of selecting a particular group of consonants from the alphabet, you would calculate it using the combination formula to determine the favorable outcomes.

In the exercise, while we are not directly calculating probability, understanding the number of possible combinations gives us the groundwork needed if a probability calculation was required in a more complex problem. Knowing how many total combinations exist allows us to set the stage for determining the likelihood of a specific event within a set of possibilities.