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- Question
36 identical books must be arranged in rows with the same number of books in each row. Each row must contain at least three books and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible ?
Options- A. 5
- B. 6
- C. 7
- D. 8
- Correct Answer
- 5
ExplanationThe following arrangements satisfy all 3 conditions.
Arrangement 1: 3 books in a row; 12 rows.
Arrangement 2: 4 books in a row; 9 rows.
Arrangement 3: 6 books in a row; 6 rows.
Arrangement 4: 9 books in a row; 4 rows.
Arrangement 5: 12 books in a row; 3 rows.Therefore, the possible arrangements are 5.
- 1. A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ?
Options- A. 210
- B. 168
- C. 1260
- D. 10!/6! Discuss
- 2. Out of seven consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?
Options- A. 25200
- B. 25000
- C. 25225
- D. 24752 Discuss
- 3. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated ?
Options- A. 15
- B. 20
- C. 5
- D. 10 Discuss
- 4. A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ? A) 11/91 B) 4/11 C) 1/11 D) 90/91
Options- A. Option A
- B. Option B
- C. Option C
- D. Option D Discuss
- 5. How many more words can be formed by using the letters of the given word 'CREATIVITY'?
Options- A. 851250
- B. 751210
- C. 362880
- D. 907200 Discuss
- 6. How many numbers of five digits can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits?
Options- A. 120
- B. 240
- C. 256
- D. 360 Discuss
- 7. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Options- A. 209
- B. 290
- C. 200
- D. 208 Discuss
- 8. How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never beings separated?
Options- A. 1440
- B. 720
- C. 360
- D. 640 Discuss
- 9. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?
Options- A. 126
- B. 120
- C. 146
- D. 156 Discuss
- 10. From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?
Options- A. 91
- B. 104
- C. 109
- D. 98 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 168
Explanation:
we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.
The captain can be selected from amongst the remaining 3 players in 3 ways.
Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.
(or)
Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 28x6 = 168 ways.
Correct Answer: 25200
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) =( ) = 210.
Number of groups, each having 3 consonants and 2 vowels =210
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 120.
Required number of words = (210 x 120) = 25200.
Correct Answer: 20
Explanation:
Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.
So, there is one way of doing it.
Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Correct Answer: Option D
Explanation:
Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14
=
ways =
= 1001
Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)
= 5+1+5 =11
P(E) =
Required probability =
Correct Answer: 907200
Explanation:
The number of letters in the given word CREATIVITY = 10
Here T & I letters are repeated
=> Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200
Correct Answer: 120
Explanation:
Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are
5 x 4 x 3 x 2 x 1 = 120.
Correct Answer: 209
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways =
= = 209.
Correct Answer: 1440
Explanation:
There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.
Correct Answer: 126
Explanation:
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows
Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.
Therefore, the total number of ways in which 8 students can travel is:
=56 + 70= 126
Correct Answer: 91
Explanation:
We first count the number of committee in which
(i). Mr. Y is a member
(ii). the ones in which he is not
Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).
We can choose 1 more in5+
=7 ways.
Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in
=84 ways.
Thus, total number of ways is 7+84= 91 ways.
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