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:
Show Hint
- \( \frac{1}{4} \)
- \( \frac{2}{3} \)
- \( \frac{1}{3} \)
- \( \frac{1}{2} \)
The Correct Option is D
Approach Solution - 1
The given word "GARDEN" contains 6 distinct letters: G, A, R, D, E, N.
The total number of possible arrangements of these 6 letters is:
\[
\text{Total arrangements} = 6! = 720
\]
The vowels in the word are A and E.
For the vowels to appear in alphabetical order (A before E), the number of valid arrangements is:
\[
\binom{6}{2} \cdot 4! = 15 \cdot 24 = 360
\]
The probability that the selected word will have vowels in alphabetical order is: \[ P = \frac{360}{720} = \frac{1}{2} \] Therefore, the probability that the selected word will NOT have vowels in alphabetical order is: \[ P(\text{Not in order}) = 1 - \frac{1}{2} = \frac{1}{2} \]
Final Answer: \( \frac{1}{2} \)Approach Solution -2
The given word is GARDEN, which contains 6 distinct letters: G, A, R, D, E, N.
Vowels: A, E.
Consonants: G, R, D, N.
Step 2: Total number of possible arrangements.
Since there are 6 distinct letters, the total number of arrangements is: \[ 6! = 720 \]
Step 3: Count the arrangements where vowels are in alphabetical order.
The vowels are A and E.
For the vowels to be in alphabetical order, A must come before E in the arrangement.
Among any two distinct letters (A and E), half of the total arrangements will have A before E, and half will have E before A.
Hence, number of arrangements where vowels are in alphabetical order: \[ \frac{6!}{2} = 360 \]
Step 4: Probability that vowels are NOT in alphabetical order.
The number of arrangements where vowels are NOT in alphabetical order is also 360 (since total = 720).
Therefore, the required probability is: \[ P = \frac{360}{720} = \frac{1}{2} \]
Final Answer:
\[ \boxed{\frac{1}{2}} \]
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