How many permutations are there of the letters of the word 'TIGER' in which the vowels should not occupy the even positions ?
36
The question asks us to find the number of permutations of the letters in the word 'TIGER' such that the vowels do not occupy the even positions. Let's break down the word and the positions.
The constraint is that the vowels (I and E) cannot be placed in the even positions (2 and 4). This implies that the vowels must be placed in the odd positions (1, 3, or 5).
We need to arrange the 5 letters (T, I, G, E, R) in the 5 positions (1, 2, 3, 4, 5) under the given condition.
The condition states that vowels cannot go into positions 2 and 4. So, the 2 vowels must be placed in the 3 odd positions (1, 3, 5).
We have 2 vowels (I, E) to be placed in 3 available odd positions (1, 3, 5). The number of ways to arrange 2 distinct items in 3 distinct positions is given by the permutation formula ${^nP_k = \frac{n!}{(n-k)!}}$, where ${n}$ is the number of positions available and ${k}$ is the number of items to be placed.
Here, ${n = 3}$ (odd positions) and ${k = 2}$ (vowels).
Number of ways to place the vowels = ${^3P_2 = \frac{3!}{(3-2)!} = \frac{3!}{1!} = 3 \times 2 \times 1 = 6}$.
After placing the 2 vowels in two of the odd positions, we have used 2 positions and have 3 positions remaining. The remaining positions are the two even positions (2 and 4) and the one odd position not occupied by a vowel. We also have 3 letters remaining, which are the 3 consonants (T, G, R).
We have 3 consonants (T, G, R) to be placed in the 3 remaining positions. The number of ways to arrange 3 distinct items in 3 distinct positions is given by ${^3P_3}$ or ${3!}$.
Number of ways to place the consonants = ${^3P_3 = 3! = 3 \times 2 \times 1 = 6}$.
To find the total number of permutations where vowels do not occupy even positions, we multiply the number of ways to place the vowels by the number of ways to place the consonants.
Total permutations = (Ways to place vowels) $\times$ (Ways to place consonants)
Total permutations = ${6 \times 6 = 36}$.
Thus, there are 36 permutations of the letters of the word 'TIGER' in which the vowels do not occupy the even positions.
| Concept | Description | Formula |
|---|---|---|
| Permutation | An arrangement of items in a specific order. | ${^nP_k = \frac{n!}{(n-k)!}}$ |
| Factorial | The product of all positive integers up to a given integer. | ${n! = n \times (n-1) \times ... \times 2 \times 1}$ |
| Constraint | A condition that limits the possible arrangements. | Handled by placing restricted items first. |
Combinatorics deals with counting, arrangement, and combination of objects. Problems often involve permutations and combinations.
In this specific problem about the word 'TIGER' and its letter arrangements with position constraints, we are dealing with permutations because the order of the letters matters for each distinct arrangement.
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