Permutations and Combinations 1 Question 12
12. Consider all possible permutations of the letters of the word ENDEANOEL.
$(2008,6 M)$
| Column I | Column II | ||
|---|---|---|---|
| A. | The number of permutations containing the word ENDEA, is |
p. | $5 !$ |
| B. | The number of permutations in which the letter E occurs in the first and the last positions, is |
q. | $2 \times 5 !$ |
| C. | The number of permutations in which none of the letters D, L, N occurs in the last five positions, is |
r. | $7 \times 5 !$ |
| D. | The number of permutations in which the letters A, E, O occur only in odd positions, is |
s. | $21 \times 5 !$ |
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Answer:
Correct Answer: 12. ($(A \rightarrow p ; B \rightarrow s ; C \rightarrow q ; D \rightarrow q)$)
Solution:
- A. If ENDEA is fixed word, then assume this as a single letter. Total number of letters $=5$
Total number of arrangements $=5$ !.
B. If $E$ is at first and last places, then total number of permutations $=7 ! / 2 !=21 \times 5$ !
C. If $D, L, N$ are not in last five positions
$\leftarrow D, L, N, N \rightarrow \leftarrow E, E, E, A, O \rightarrow$
Total number of permutations $=\frac{4 !}{2 !} \times \frac{5 !}{3 !}=2 \times 5$ !
D. Total number of odd positions $=5$
Permutations of AEEEO are $\frac{5 !}{3 !}$.
Total number of even positions $=4$
$\therefore$ Number of permutations of N, N, D, L $=\frac{4 !}{2 !}$
$\Rightarrow$ Total number of permutations $=\frac{5 !}{3 !} \times \frac{4 !}{2 !}=2 \times 5$ !
