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- Question
From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ?
Options- A. 7600
- B. 7200
- C. 6400
- D. 3600
- Correct Answer
- 7200
ExplanationFrom 5 consonants, 3 consonants can be selected in ways.
From 4 vowels, 2 vowels can be selected in ways.
Now with every selection, number of ways of arranging 5 letters is ways.
Total number of words =
= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200
- 1. How many arrangements of the letters of the word ?BENGALI? can be made if the vowels are to occupy only odd places.
Options- A. 720
- B. 576
- C. 567
- D. 625 Discuss
- 2. Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.
Options- A. 340
- B. 370
- C. 320
- D. 330 Discuss
- 3. In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels always come together?
Options- A. 720
- B. 1440
- C. 1800
- D. 3600 Discuss
- 4. The number of ways in which 8 distinct toys can be distributed among 5 children?
Options- A. 5P8
- B. 5^8
- C. 8P5
- D. 8^5 Discuss
- 5. How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ?
Options- A. 521
- B. 720
- C. 420
- D. 225 Discuss
- 6. There are three rooms in a Hotel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?
Options- A. 105
- B. 7! x 6!
- C. 7!/5!
- D. 420 Discuss
- 7. In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?
Options- A. 186
- B. 144
- C. 136
- D. 120 Discuss
- 8. Nine different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, how many such words can be formed which have at least one letter repeated ?
Options- A. 43929
- B. 59049
- C. 15120
- D. 0 Discuss
- 9. To fill 8 vacancies there are 15 candidates of which 5 are from ST. If 3 of the vacancies are reserved for ST candidates while the rest are open to all, Find the number of ways in which the selection can be done ?
Options- A. 7920
- B. 74841
- C. 14874
- D. 10213 Discuss
- 10. The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is
Options- A. 7! x 7!
- B. 7! x 6!
- C. 6! x 6!
- D. 7! x 5! Discuss
Permutation and Combination problems
Search Results
Correct Answer: 576
Explanation:
There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.
There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in ways and 4 constants can be arranged in ways.
Number of words = x = 24 x 24 = 576
Correct Answer: 330
Explanation:
Here the order of choosing the elements doesn?t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in 11 ways = 330 ways
Correct Answer: 1440
Explanation:
Given word is THERAPY.
Number of letters in the given word = 7
Number of vowels in the given word = 2 = A & E
Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is
6! x 2! = 720 x 2 = 1440.
Correct Answer: 5^8
Explanation:
As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are ways to distribute the toys.
Hence, it is and not .
Correct Answer: 720
Explanation:
Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240
Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.
Correct Answer: 105
Explanation:
Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,
Then, 7C1 x 6C2 x 4C4
= 7 x 15 x 1 = 105
Correct Answer: 120
Explanation:
In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.
There are 6 letters in thegiven word.
First arrange 3 vowels.
This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)
Remaining 3 letters can be placed in 3 places = 3! ways
Total number of possible ways of arranging letters of OLIVER = 3! x ways = 6x5x4 = 120 ways.
Correct Answer: 43929
Explanation:
Number of words with 5 letters from given 9 alphabets formed =
Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is =
Number of words can be formed which have at least one letter repeated =
= 59049 - 15120
= 43929
Correct Answer: 7920
Explanation:
ST candidates vacancies can be filled by ways = 10
Remaining vacancies are 5 that are to be filled by 12
=> = (12x11x10x9x8)/(5x4x3x2x1) = 792
Total number of filling the vacancies = 10 x 792 = 7920
Correct Answer: 7! x 6!
Explanation:
The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in ways. 7 teachers can sit in (7-1)! ways.
Required no.of ways is =
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