Question

How many different words can be formed using all the letters of the word "SPECIAL", so that the consonants always in the odd positions? (1) 112 (2) 72 (3) 24 (4) 144

Short answer

Answer: 144

Step-by-step solution

Identify the consonants and vowels

The word "SPECIAL" has the following consonants: S, P, C, and L, and the following vowels: E, I, and A.

Place consonants in odd positions

Since there are 4 consonants, they can occupy the odd positions. There are a total of 4 odd positions in a 7-letter word (1, 3, 5, and 7). So, the number of ways to place the 4 consonants in the odd positions is the number of permutations of 4 elements, which is computed as 4! = 4 × 3 × 2 × 1 = 24.

Place vowels in even positions

The remaining 3 positions (2, 4, and 6) are for the vowels. There are 3! = 3 × 2 × 1 = 6 ways to place the vowels, since there are 3 vowels, and they can be arranged among themselves in 3! ways.

Calculate the total number of words

The total number of words is the product of the number of ways to place consonants and the number of ways to place vowels. Therefore, the total number of different words that can be formed is 24 (consonants) × 6 (vowels) = 144. Hence, the correct answer is option (4) 144.

Key concepts

Permutations and Combinations

Permutations and combinations are fundamental concepts in combinatorial mathematics, often used to solve problems related to counting. A permutation refers to the arrangement of a set of objects in a specific order. The key aspect of permutations is that the order matters; 'ABC' is a different permutation than 'BCA'.

In contrast, a combination represents a selection of objects where the order is not important. For instance, the combination of letters 'ABC' represents the same selection as 'CBA'. In our problem, we're dealing only with permutations because the position of each letter matters to form different words.

Factorial Notation

Factorial notation is a mathematical operation that multiplies a number by every number below it, down to 1. It's symbolized using an exclamation point (!). For example, 5! (read as 'five factorial') is calculated as 5 × 4 × 3 × 2 × 1. Factorials are especially useful in permutations and combinations because they count the number of ways a set number of items can be arranged. In the permutation of consonants 'SPCL', 4! is used to determine the different arrangements.

Arranging Letters

Arranging letters of a word involves finding different permutations, especially when there are several conditioning rules, such as in our exercise where consonants and vowels must occupy certain positions. This is common in word problems involving permutations. To systematically solve these, one should separate the distinct elements (like consonants and vowels) and consider their individual arrangements before combining them to find the total possible arrangements. Our step-by-step solution uses this approach effectively.

Odd and Even Positions

In permutation problems, it can be important to consider the odd and even positions that elements can occupy. Odd positions are 1st, 3rd, 5th, and so on, while even positions are 2nd, 4th, 6th, etc. In the context of our problem, this translates to placing consonants in the 1st, 3rd, 5th, and 7th positions, and vowels in the 2nd, 4th, and 6th positions. The constraints imposed by positions lead to different permutations and are a classic constraint in combinatorial problems.