Discussion
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- Question
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
- Correct Answer
- 4320
ExplanationLet '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.
- 1. In how many different ways could couples be picked from 6 men and 9 women ?
Options- A. 26
- B. 54
- C. 52
- D. 28 Discuss
- 2. 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 Discuss
- 3. If a+b+c =21 what is the total number of positive integral solutions?
Options- A. 109
- B. 190
- C. 901
- D. 910 Discuss
- 4. A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?
Options- A. 569
- B. 729
- C. 625
- D. 769 Discuss
- 5. If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?
Options- A. 720
- B. 360
- C. 240
- D. 740 Discuss
- 6. 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 Discuss
- 7. 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 Discuss
- 8. 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
- 9. 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 Discuss
- 10. 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 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 54
Explanation:
Number of mens = 8
Number of womens = 5
Different ways could couples be picked = = 9 x 6 = 54 ways.
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.
Correct Answer: 190
Explanation:
Number of positive integral solutions = = = 190
Correct Answer: 729
Explanation:
Choose 5 starters from a team of 12 players. Order is not important.
= 729
Correct Answer: 360
Explanation:
We have to find number of permutations of 4 objects out of 6 objects.
This number is = 360
Therefore, cards can be sent in 360 ways.
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.
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.
Correct Answer: 34
Explanation:
7 × 5 = 35
35 ? 1 = 34
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
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
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