Permutation and Combination problems
- 1. If two cards are taken one after another without replacing from a pack of 52 cards. What is the probability for the two cards be Ace ?
Options- A. 51/1221
- B. 42/221
- C. 1/221
- D. 52/1245 Also asked in: CAT, Bank Exams, AIEEE, Bank PO, Bank Clerk, Analyst
Correct Answer: 1/221
Explanation:
Total Combination of getting a card from 52 cards =
Because there is no replacement, so number of cards after getting first card= 51
Now, Combination of getting an another card=
Total combination of getting 2 cards from 52 cards without replacement= (
)
There are total 4 Ace in stack. Combination of getting 1 Ace is =
Because there is no replacement, So number of cards after getting first Ace = 3
Combination of getting an another Ace =
Total Combination of getting 2 Ace without replacement=
Now,Probability of getting 2 cards which are Ace = = 1/221.
- 2. In how many different ways could couples be picked from 6 men and 9 women ?
Options- A. 26
- B. 54
- C. 52
- D. 28 Also asked in: GATE, CAT, Bank Exams, AIEEE, Bank PO, Bank Clerk
Correct Answer: 54
Explanation:
Number of mens = 8
Number of womens = 5
Different ways could couples be picked = = 9 x 6 = 54 ways.
- 3. The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?
Options- A. 2580
- B. 3687
- C. 4320
- D. 5460 Also asked in: AIEEE, Bank Exams, CAT, GATE, Bank Clerk, Bank PO
Correct Answer: 4320
Explanation:
Let 'Y' be the youngest player.
The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.
The required number of ways = 3(3!) = 4320.
- 4. In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?
Options- A. 720
- B. 360
- C. 120
- D. 24 Also asked in: AIEEE, Bank Exams, CAT, GATE, Analyst, Bank Clerk, Bank PO
Correct Answer: 24
Explanation:
Given word is SCOOTY
ATQ,
Except S & Y number of letters are 4(C 2O's T)
Hence, required number of arrangements = 4!/2! x 2! = 4!
= 4 x 3 x 2
= 24 ways.
- 5. A box contains 2 blue balls, 3 green balls and 4 yellow balls. In how many ways can 3 balls be drawn from the box, if at least one green ball is to be included in the draw?
Options- A. 48
- B. 24
- C. 64
- D. 32 Also asked in: AIEEE, Bank Exams, CAT, Analyst, Bank Clerk, Bank PO
Correct Answer: 64
Explanation:
Total number of balls = 2 + 3 + 4 = 9
Total number of ways 3 balls can be drawn from 9 = 9C3
No green ball is drawn = 9 - 3 = 6 = 6C3
Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn
= 9C3 - 6C3
= 9x8x7/3x2 - 6x5x4/3x2
= 84 - 20
= 64 ways.
- 6. Out of 6 green ball, 4 blue ball, in how many ways we selectone or more balls ?
Options- A. 42
- B. 34
- C. 31
- D. 22 Discuss
Correct Answer: 34
Explanation:
7 × 5 = 35
35 ? 1 = 34
- 7. If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is
Options- A. 9!
- B. 9!/2
- C. 10!
- D. 10!/2 Also asked in: CAT, Bank Exams, AIEEE, Bank PO, Bank Clerk
Correct Answer: 10!/2
Explanation:
As A1 speaks always after A2, they can speak only in 1st to 9th places and
A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place
A2 can speak in 9 places the remaining
A3, A4, A5,...A10 has no restriction. So, they can speak in 9.8! ways. i.e
when A2 speaks in the first place, the number of ways they can speak is 9.8!.
When A2 speaks in second place, the number of ways they can speak is 8.8!.
When A2 speaks in third place, the number of ways they can speak is 7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!
Therefore,Total Number of ways they can speak = (9+8+7+6+5+4+3+2+1) 8! = = 10!/2
- 8. In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together?
Options- A. 360
- B. 720
- C. 120
- D. 840 Also asked in: AIEEE, Bank Exams, CAT, Bank Clerk
Correct Answer: 360
Explanation:
CAPITAL = 7
Vowels = 3 (A, I, A)
Consonants = (C, P, T, L)
5 letters which can be arranged in
Vowels A,I =
No.of arrangements = 5! x =360
- 9. How many arrangements of the letters of the word ?BENGALI? can be made if the vowels are never together.
Options- A. 120
- B. 640
- C. 720
- D. 540 Discuss
Correct Answer: 720
Explanation:
There are 7 letters in the word ?Bengali of these 3 are vowels and 4 consonants.
Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.
Total number of words = 5! x 3!= 120 x 6 = 720
- 10. There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf ?
Options- A. 26100
- B. 21600
- C. 24000
- D. 36000 Discuss
Correct Answer: 21600
Explanation:
4 novels can be selected out of 5 in
ways.
2 biographies can be selected out of 4 in
ways.
Number of ways of arranging novels and biographies =
= 30
After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.
By the Counting Principle, the total number of arrangements = 30 x 720 = 21600
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