Question ID: #479
Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is:
- (1) $\frac{1}{4}$
- (2) $\frac{2}{3}$
- (3) $\frac{1}{3}$
- (4) $\frac{1}{2}$
Solution:
The word is GARDEN.
Total letters = 6 (G, A, R, D, E, N).
Vowels are A and E.
Total number of words formed ($n(S)$) = $6!$.
We need to consider the relative order of the vowels A and E.
In any arrangement, there are only two possibilities for the relative order of A and E:
1. A comes before E (Alphabetical order).
2. E comes before A (Not in Alphabetical order).
Due to symmetry, the number of words where A comes before E is equal to the number of words where E comes before A.
Let $E_1$ be the event that vowels are in alphabetical order.
Let $E_2$ be the event that vowels are NOT in alphabetical order.
$P(E_1) = P(E_2) = \frac{1}{2}$.
Alternatively, by calculation:
Number of ways to arrange vowels A and E in alphabetical order:
Select 2 positions for vowels out of 6: ${^6}C_2$.
Place A and E in these positions in 1 way (A then E).
Arrange the remaining 4 consonants in the remaining 4 positions: $4!$.
Favorable cases for alphabetical order = ${^6}C_2 \times 1 \times 4!$.
$$= 15 \times 24 = 360$$
Total cases = $6! = 720$.
$$P(\text{Alphabetical Order}) = \frac{360}{720} = \frac{1}{2}$$
Required Probability ($P(\text{NOT Alphabetical})$) = $1 – P(\text{Alphabetical Order})$
$$= 1 – \frac{1}{2} = \frac{1}{2}$$
Ans. (4)
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