வெல்லுங்கள் CSAT 2025 - 5: கொள்குறி வினா - விடைகள் - Permutations & Combinations

Questions & Answers (Mutiple Choice)
கொள்குறி வினா- விடைகள்

Permutations & Combinations

1. Ramesh has 5 pairs of white, 4 pairs of black and 3 pairs of green socks in a box. If he randomly picks socks one by one (without replacement) from the box to get a black pair to wear, find the maximum number of attempts he has to make.
a. 22 b. 20 c. 18 d. 16

2. Using all the digits of the Kaprekar constant 6174 without repetition, we can form 24 distinct 4-digit numbers. Out of these 24 numbers, how many of them are exactly divisible by 18?
a. 24 b. 12 c. 18 d. 16

3. A gentleman decided to host a dinner to 5 friends to be selected out of ten friends. Find the number of ways forming the party of 5 given that two of the friends will not attend the party together.
a. 252 b. 56 c. 308 d. 196

4. Find the sum of all 4-digit numbers that can be formed by using the digits
1, 2, 3 and 4 without repetition of the digits.
a. 55550 b. 60000 c. 66660 d. 6660

5. Eleven friends meet at a party. Each handshakes one another once. How many handshakes might have done at the party?
a. 132 b. 66 c. 110 d. 55

6. How many diagonals in a Decagon?
a. 35 b. 45 c. 50 d. 60

7. A child is in a position to select from
4 chocolates, 5 sugar candies and 3 fruits. In how many ways the child make the selection?
a. 120 b. 119 c. 60 d. 59

8. A group has 4 boys and 5 girls. In how many ways can they seated in a row so that all boys do not sit together?
a. 6! b. 9! c. 9! - 6! d. 9! - 6!4!

9. A student is to answer 10 out of 13 questions in an examination such that he must choose atleast 4 from first 5 questions. The number of choices available to him is
a. 140 b. 280 c. 196 d. 346

10. How many 3-digit numbers can be formed with the digits 0, 2, 4, 6, 7 and 8
if the repetition of digits not allowed?
a. 100 b. 120 c. 180 d. 160

11. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
a. 120 b. 720 c. 1440 d. 5040

12. An organization has three committees. Only two persons are members of all committees but every pair of committees have three members in common. What is the least possible members in total of all the three committees?
a. 8 b. 7 c. 6 d. 5

13. A dictionary is made up of the words (whether they have meaning or not) that can be done by arranging the letters of the word ALTER. What is the 50th word if words are printed in the same order as that of ordinary dictionary?
a. LAERT b. LARTE c. LAETR d. LATRE

14. There are 6 locks and 6 keys. What is the maximum number of trials one should make to establish the correspondence between the locks and keys?
a. 20 b. 21 c. 24 d. 30

15. There are 7 non - collinear points. The number of triangles can be drawn by joining these points is
a. 21 b. 35 c. 14 d. 7

Answers

1. c 2. b 3. d 4.c 5. d

6. a 7. b 8. d 9. c 10. a

11. b 12.d 13.c 14. a 15. b

Explanatory Answers

1. Total number socks in the box
= (5+4+3)×2 = 24.
Number of black socks in the box
= 4×2 =8
Maximum attempts that no black sock pick is 24 - 8 =16

Hence the Maximum attempts that two black socks pick is = 16 + 2 = 18.

2. Since the digital extract of 6174 is 9
(6+1+7+4 =18 -> 1+8 -> 9), all 24 distinct 4-digit numbers are divisible by 9.
We know the even numbers whose digital extract is 9 are always divisible by 18. So even numbers using all the digits of 6174 without repetition are divisible by 18.
Here numbers ending with 4 or 6 are even.
Fix the unit digit either 4 or 6 .
Thousandth place can be filled by in 3 ways. Hundredth place can be filled by 2 in ways.Tenth place can be filled by only in one way.
Hence the number of such possible even numbers = 3×2×1×2 =12.

3. Out of 10, to select 5,
the number of ways = 10c5
Two of the friends always together means to make them invited.
The number of ways to select the other three from 8 is 8c3.
Required number of ways = 10c5 - 8c3
10c5 = (10×9×8×7×6)/(1×2×3×4×5)= 252
8c3 = (8×7×6)/(1×2×3) = 56
Hence 252 - 56 = 196.

4. The number of 4-digit numbers that can be formed with 1, 2, 3 and 4 without repetition = 4! = 24.
If 4 occurs at the unit place, then the remaining digits can be arranged in
3p3 ways i.e. 3×2×1 = 6 ways.
Similarly other three digits occur at the unit places in 6 ways.
According to unit place, the sum of all 24 numbers = 6(1+2+3+4) = 60
Likewise 10th place etc.,
Hence the sum of all 24 numbers
= 60000 + 6000 + 600 + 60 = 66660

5. Number of handshakes = 11c2
= (11×10)/1×2 = 55

6. The number of diagonals of a
' n ' sided polygon is
nC2 - n = [n(n-1)/2] - n = [n(n - 3)]/2.
Here n = 10 ;
So Required number of diagonals
= (10×7)/2 = 35

7. The total number of ways including non selection is
(4+1)(5+1)(3+1) = 120
We bear in mind that
Out of 4 chocolates, the selection will be
Nil, 1, 2, 3 and 4. Total number of ways (4+1) = 5.
Hence the required number of ways
= 120 - 1 = 119.

8. 9 persons can be arranged to sit in a row in 9! ways.
If all boys are together, then 4 boys can be taken as one person. But they can be arranged to sit in 4! ways.
Hence 4 boys and 5 girls can be seated in 6! 4! if all boys seated together.
Hence the number of ways can they seated in a row so that all boys do not sit together is 9! - 6!4!

9. Atleast 4 means, he can select either 4 or 5 questions from first 5 questions.
First 5 questions and the rest 8 questions.
So the required number of choices is
5c4 × 8c6 + 5c5 × 8c5
= 5×28 + 1×56 = 196

10. The digit 0 cannot be the hundredths digit of a 3-digit number.
So hundredth place can be filled by any one of the other 5 digits.
But tenth place can be filled by any one of the 5 digits.
And unit place can be filled by any one of the remaining 4 digits.
The number of 3-digit numbers csn be formed = 5 × 5 × 4 = 100

11. The word 3 vowels and 4 consonants. Since all the letters are different, the totsl number of words formed = 7!
Take all the 3 vowels as one letter. But the vowels themselves can be arranged in 3! ways. Then the number of words formed with vowels together and 4 consonants is
(4+1)! 3! = 5! 3! = 120×6 = 720.

12.

Each Committee must have 4 members.
The least possible members in total of all the three committees is 5.

13. Alphabetically the letters in the word ALERT are A, E, L, R, T
Starting with A, there must be
4! = 24 words. Similarly Starting with E, there must be 4! = 24 words.
49th word starts with L. And the word is
LAERT. So 50th word is LAETR.

14. Lock 1 - Correct key can be identified in 6 trials. Similarly Lock 2 - in 5 trials;
Lock 3 - in 4 trials; Lock 4 - in 3 trials;
Lock 5 - in 2 trials;
Lock 6 - correct fit the last one key
The total number of trials =
6 + 5 + 4 +3 +2 = 20

15. Out of 7, three can be selected in
7c3 ways.
So (7×6×5)/(1×2×3) = 35.

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