How many words can be formed by using all the letters of the word, 'ALLAHABAD' ?
|
| A.
| 3780
|
|
| B.
| 1890
|
|
| C.
| 7560
|
|
| D.
| 2520
|
Answer & Explanation
Answer: Option C
Explanation:
The word 'ALLAHABAD' contains 9 letters, namely 4A, 2L, 1H, 1B and 1D.
$$\therefore$$ Required number of words = $$\frac{9 !}{(4 !) (2 !) (1 !)(1 !)(1 !)}$$ = 7560.
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the
vowels always come together ?
|
| A.
| 360
|
|
| B.
| 480
|
|
| C.
| 720
|
|
| D.
| 5040
|
Answer & Explanation
Answer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LDNG (EAI).
Now, 5 letters can be arranged in 5 ! = 120 ways
The vowels (EAI) can be arranged among themselves in 3 ! = 6 ways.
$$\therefore$$
Required number of ways = (120 * 6) = 720.
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together ?
|
| A.
| 10080
|
|
| B.
| 4989600
|
|
| C.
| 120960
|
|
| D.
| None of these
|
Answer & Explanation
Answer: Option C
Explanation:
In the word 'MATHEMATICS' we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = $$\frac{8 !}{(2 !) (2 !)}$$ = 10080.
Now, AEAI has
4 Letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = $$\frac{4 !}{2 !}$$ = 12.
$$\therefore$$ Required number of words = (10080 * 12) = 120960.
In how many different ways can the letters of the word 'DETAIL be arranged in such a way that the vowels occupy only the odd positions ?
Answer & Explanation
Answer: Option C
Explanation:
There are 6 letters in the given word out of which there are 3 vowels and 3 consonants. Let us mark these positions as under :
(1)(2)(3)(4)(5)(6)
Now, 3 vowels can be placed at any of the three places out of 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3 ! =
6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3 ! = 6 !.
Total number of ways = (6 * 6) = 36.
In how many different ways can the letters of the word 'AUCTION' be arranged in such a way that the vowels always come together ?
|
| A.
| 30
|
|
| B.
| 48
|
|
| C.
| 144
|
|
| D.
| 576
|
Answer & Explanation
Answer: Option D
Explanation:
The word 'AUCTION' has 7 different letters.
When the vowels AUIO are always together, they can be supposed to form one letter.
Then, we have to arrange the letters CTN (AUIO).
Now, 4 letters can be arranged in 4 ! = 24 ways.
The vowels (AUIO) can be arranged among themselves in 4 ! = 24 ways.
$$\therefore$$ Required number of ways = (24 * 24) = 576.
In how many different ways can the letters of the word 'JUDGE' be arranged in such a way that the vowels always come together ?
|
| A.
| 48
|
|
| B.
| 120
|
|
| C.
| 124
|
|
| D.
| 160
|
Answer & Explanation
Answer: Option A
Explanation:
The word 'JUDGE' has 5 different letters.
When the vowels UE are always together, they can be supposed to form one letter.
Then, we have to arrange the letters JDG (UE).
Now, 4 letters can be arranged in 4 ! = 24 ways.
The vowels (UE) can be arranged among themselves in 2 ! = 2 ways.
$$\therefore$$
Required number of ways = (24 * 2) = 48.
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together ?
|
| A.
| 810
|
|
| B.
| 1440
|
|
| C.
| 2880
|
|
| D.
| 50400
|
Answer & Explanation
Answer: Option D
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 letters of which R occurs 2 times and the rest are different.
Number of ways of arranging these letters = $$\frac{7 !}{2 !}$$ = 2520.
Now, 5 vowels in which O occurs 3 times and the rest
dre different, can be arranged in $$\frac{5 !}{3 !}$$ = 20 ways.
$$\therefore$$ Required number of ways = (2520 * 20) = 50400.
In how many different ways can the letters of the word 'BANKING' be arranged so that the vowels always come together ?
|
| A.
| 120
|
|
| B.
| 240
|
|
| C.
| 360
|
|
| D.
| 720
|
Answer & Explanation
Answer: Option D
Explanation:
In the word 'BANKING', we treat the two vowels AI as one letter. Thus, we have BNKNG (AI).
This has 6 letters of which N occurs 2 times and the rest are different.
Number of ways of arranging these letters = $$\frac{6 !}{(2 !)(1 !)(1 !)(1 !)(1 !)}$$ = 360.
Now, 2 vowels AI can be arranged in 2 ! = 2
ways.
$$\therefore$$ Required number of ways = (360 * 2) = 720.
In how many different ways can the letters of the word 'MACHINE' be arranged so that the vowels may occupy only the odd positions ?
|
| A.
| 210
|
|
| B.
| 576
|
|
| C.
| 144
|
|
| D.
| 1728
|
Answer & Explanation
Answer: Option B
Explanation:
There are 7 letters in the given word, out of .which there are 3 vowels and 4 consonants. Let us mark the positions to be filled up as follows :
(1)(2)(3)(4)(5)(6)(7)
Now, 3 vowels can be placed at any of the three places, out of the four marked 1, 3, 5,7.
Number of ways of arranging the vowels =
4P3 = (4 *3 * 2) = 24.
Also, the 4 consonants at the remaining 4 positions may be arranged in = 4P4 = 4 ! = 24 ways.
$$\therefore$$ Required number of ways = (24 * 24) = 576.
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women ?
|
| A.
| 266
|
|
| B.
| 5040
|
|
| C.
| 11760
|
|
| D.
| 86400
|
Answer & Explanation
Answer: Option C
Explanation:
Required number of ways = (8C5 * 10C6)
= 8C3 * 10C4 = $$(\frac{8 * 7 * 6}{3 * 2 * 1} * \frac{10 * 9 * 8 * 7}{4 * 3 * 2 * 1})$$ = 11760.
The number of ways the word TRAINER can be arranged so that the vowels always come together are 360.
