Discussion
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- Question
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
- Correct Answer
- 43929
ExplanationNumber 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
- 1. 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
- 2. 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
- 3. 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 Discuss
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?
Options- A. 180
- B. 220
- C. 240
- D. 160 Discuss
- 9. How many six digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8, and 9 (repetition not allowed)?
Options- A. 8640
- B. 720
- C. 3620
- D. 4512 Discuss
- 10. On the occasion of New Year, each student of a class sends greeting cards to the others. If there are 21 students in the class, what is the total number of greeting cards exchanged by the students?
Options- A. 380
- B. 420
- C. 441
- D. 400 Discuss
Permutation and Combination problems
Search Results
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: 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: 7200
Explanation:
From 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
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: 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 =
Correct Answer: 180
Explanation:
Number of ways of choosing 2 black pens from 5 black pens in ways.
Number of ways of choosing 2 white pens from 3 white pens in ways.
Number of ways of choosing 2 red pens from 4 red pens in ways.
By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.
Correct Answer: 8640
Correct Answer: 420
Explanation:
Given total number of students in the class = 21
So each student will have 20 greeting cards to be send or receive (21 - 1(himself))
Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.
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