Aptitude questions and answers on Permutations

Aptitude questions and answer from permutations

1. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A. 1260

B. 210

C. 10C6 * 6!

D. 10C5 * 6

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, the total way the selection can be made is 210*6 = 1260.

2. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A. 24400

B. 21300

C. 210

D. 25200

Explanation:

Number of ways of selecting 3 consonants from 7= 7C3
Number of ways of selecting 2 vowels from 4= 4C2
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
=(7×6×5/3×2×1)×(4×3/2×1)=210
It means we can have 210 groups where each group contains a total of 5 letters (3 consonants and 2 vowels).
The number of ways of arranging 5 letters among themselves=5!=5×4×3×2×1=120
Hence, the required number of ways=210×120=25200

3. A boy has nine pairs of trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?

A. 21

B. 12

C. 9

D. 108

Explanation:

The boy can select one pair of trousers in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.

4. Find the sum of all the 4 digit numbers that can be formed with the digits 3, 4, 5 and 6

A. 119988

B. 11988

C. 191988

D. None of these

Explanation:

No. of Digits = 4 All are distinct; They can be arranged in 4! = 24 ways
Each of the digits 3, 4, 5, and 6 occur at unit place = 3! Ways = 6 ways.
Thus there will be 6 numbers ending with 3, 4, 5 and 6 each. So the sum of the digits at unit's place = 6(3 + 4 + 5 + 6) =108
The sum of numbers = 108 × 103 + 108 × 102 + 108 × 101 + 108 × 100 = 119988

5. 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. 720

B. 520

C. 700

D. 750

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 LNDG (EAI).
Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.

6. How many combinations of students are possible if the group is to consist of exactly 3 freshmen?

A. 5000

B. 4550

C. 4000

D. 3550

Explanation:

Here we need the number of possible combinations of 3 out of 5 freshmen,5C3, and the number of possible combinations of 3 out of the 15 sophomores and juniors,
15C3. Note that we want 3 freshmen and 3 students from the other classes. Therefore, we multiply the number of possible groups of 3 of the 5 freshmen times the number of possible groups of 3 of the 15 students from the other classes. 5C3×15C3=4,550.

7. Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?

A. 10080

B. 4569

C. 9800

D. 2478

Explanation:

The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice. Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.

8. A five-digit number is formed using digits 1, 3, 5, 7, and 9 without repeating any one of them. What is the sum of all such possible numbers?

A. 6666600

B. 6666660

C. 6666666

D. None of these

Explanation:

If we fix 1 at the unit’s place, the other digits can be arranged in 4! = 24 ways.
So, there are 24 numbers which have 1 at the unit’s place and some of these numbers is 24.
Similarly, there will be 24 numbers each with 3, 5, 7, and 9 at the unit’s place.
So, the sum of all the numbers at the unit’s place will be:
24 + 24 × 3 + 24 × 5 + 24 × 7 + 24 × 9 = 24 + 72 + 120 + 168 + 216 = 600.
Similarly, the sum of the digits at the ten’s place, hundred’s place,
thousand’s place, and ten thousand’s place will be 600.
So, the sum of all numbers = 600 (1 + 10 +100 +1,000 +10,000) = 6666600.

9. An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can
however potentially create confusion when reading upsidedown-for example, code 91 may appear as 16. How many codes are there for which no such confusion can
arise?

A. 80

B. 78

C. 71

D. 69

Explanation:

The available digits are 0,1,2, …9.
The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digits repetition not allowed).
Thus, the code can be made in 9 × 9 = 81 ways.
Now there are only 4 digits 1, 6, 8, 9 which can create confusion.
Hence, the total number of codes that create confusion is = 4 × 3 = 12.
Out of these 12 codes, 69 and 96 will not create confusion.
Hence, in total 12 – 2 = 10 codes will create confusion.
Hence, the total codes without confusion are 81 – 10 = 71.

10. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A. 1260

B. 1400

C. 1250

D. 1600

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways. Now, the captain can be selected from these 6 players in 6 ways.
Therefore, the total way the selection can be made is 210×6= 1260.

11. How many words can be formed by using all letters of TIHAR?

A. 100

B. 120

C. 140

D. 160

Explanation:

The first thing to understand in this question is that it is a permutation question.
Total number of words = 5
Required number = 5P5=5!=5*4*3*2*1=120

12. How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS' if repetition of letters is not allowed?

A. 720

B. 420

C. None of these

D. 5040

Explanation:

The word 'LOGARITHMS' has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= 10P3
= 10×9×8 = 720

13. In how many different ways can the letters of the word 'DETAIL' be arranged such that the vowels must occupy only the odd positions?

A. None of these

B. 64

C. 120

D. 36

Explanation:

The word 'DETAIL' has 6 letters which have 3 vowels (EAI) and 3 consonants(DTL)
The 3 vowels(EAI) must occupy only the odd positions. Let's mark the positions as (1) (2) (3) (4) (5) (6). Now, the 3 vowels should only occupy the 3 positions marked as (1),(3), and (5) in any order.
Hence, a number of ways to arrange these vowels
= 3P3
= 3!=3×2×1=6
Now we have 3 consonants(DTL) which can be arranged in the remaining 3 positions in any order. Hence, a number of ways to arrange these consonants
= 3P3
=3!=3×2×1=6
Total number of ways
= number of ways to arrange the vowels × number of ways to arrange the consonants
=6×6=36

14. 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

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
A number of ways arranging these letters =7!= 2520.2!
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in 5!/3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.

15. Find the sum of all the 4 digit numbers that can be formed with the digits 3, 4, 5 and 6

A. 119988

B. 11988

C. 191988

D. None of these

Explanation:

No. of Digits = 4 All are distinct; They can be arranged in 4! = 24 ways
Each of the digits 3, 4, 5, and 6 occur at unit place = 3! Ways = 6 ways.
Thus there will be 6 numbers ending with 3, 4, 5 and 6 each. So the sum of the digits at unit's place = 6(3 + 4 + 5 + 6) =108
The sum of numbers = 108 × 103 + 108 × 102 + 108 × 101 + 108 × 100 = 119988

16. In a cricket championship, there are 21 matches. If each team plays one match with every other team, the number of teams is

A. 7

B. 9

C. 10

D. None of these

Explanation:

Let n be the number of teams.
nC2 = 21
(n(n-1)/2) = 21
n(n-1) = 42
n = 7

17. A committee has 5 men and 6 women. What is the number of ways of selecting 2 men and 3 women from the given committee?

A. 150

B. 200

C. 250

D. 300

Explanation:

The number of ways to select two men and three women = 5C2 * 6C3
= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)
= 200

18. There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed so that a passenger can travel from any station to any other
station?

A. 200

B. 190

C. 95

D.380

Explanation:

The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in 20P2 ways.
20P2 = 20 * 19 = 380.

19. When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

A. 155

B. 620

C. 671

D. 625

Explanation: C

When 4 dice are rolled simultaneously, there will be a total of 64 = 1296 outcomes.
The number of outcomes in which none of the 4 dice show 3 will be 54 = 625 outcomes.
Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

20. Four dice are rolled simultaneously. What is the number of possible outcomes in which at least one of the die shows 6?

A. 6! / 4!

B. 625

C. 671

D. 1296

Explanation:

When 4 dice are rolled simultaneously, there are 64 = 1296 outcomes. The converse of what is asked in the question is that none of the dice show '6'. That is all four dice show any of the other 5 numbers. That is possible in 54 = 625 outcomes. Therefore, in 1296 - 625 = 671 outcomes at least one of the dice will show 6.

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