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- Question
There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?
Options- A. 1170
- B. 1200
- C. 720
- D. 360
- Correct Answer
- 1170
ExplanationPossibilities Bowlers Batsmen Number of ways
6 9
1 4 7
2 5 6
3 6 5
= 15 x 36 = 540
= 6 x 84 = 504
= 1 x 126 = 126
Total = 1170
- 1. There are 7 non-collinear points. How many triangles can be drawn by joining these points?
Options- A. 10
- B. 30
- C. 35
- D. 60 Discuss
- 2. How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?
Options- A. 59
- B. 56
- C. 64
- D. 55 Discuss
- 3. In How many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?
Options- A. 131
- B. 191
- C. 68
- D. 3720 Discuss
- 4. Four ladies A, B, C and D and four gentlemen E, F, G and H are sitting in a circle round a table facing each other. Directions: (1) No two ladies or two gentlemen are sitting side by side. (2) C, who is sitting between G and E is facing D. (3) F is between D and A and is facing G. (4) H is to the right of B. Question: 1. Who are immediate neighbours of B? 2. E is facing whom?
Options- A. G & H , H
- B. F & H , B
- C. E & F , F
- D. E & H , G Discuss
- 5. If nC10=nC12 then,find n.
Options- A. 10
- B. 12
- C. 22
- D. 24 Discuss
- 6. Find the value of 'n' for which the nth term of two AP'S: 15,12,9.... and -15,-13,-11...... are equal?
Options- A. n = 2
- B. n = 5
- C. n = 29/5
- D. n = 1 Discuss
- 7. In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?
Options- A. 18000
- B. 17280
- C. 17829
- D. 18270 Discuss
- 8. In how many different ways can the letters of the word 'POVERTY' be arranged ?
Options- A. 2520
- B. 5040
- C. 1260
- D. None Discuss
- 9. In how many different ways can the letters of the word 'ABYSMAL' be arranged ?
Options- A. 5040
- B. 3650
- C. 4150
- D. 2520 Discuss
- 10. Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?
Options- A. 2400
- B. 6400
- C. 3600
- D. 7200 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 35
Explanation:
A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points
The number of triangles formed = = 35
Correct Answer: 59
Explanation:
The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or have both the letters to be the same.
Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.
Total number of possibilities = 56 + 3 = 59
Correct Answer: 191
Explanation:
Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.
Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191
Correct Answer: G & H , H
Correct Answer: 22
Explanation:
Using, we get
n ? 10 = 12
or, n = 12 + 10 = 22
Correct Answer: n = 29/5
Explanation:
Given are the two AP'S:
15,12,9.... in which a=15, d=-3.............(1)
-15,-13,-11..... in which a'=-15 ,d'=2.....(2)
now using the nth term's formula,we get
a+(n-1)d = a'+(n-1)d'
substituting the value obtained in eq. 1 and 2,
15+(n-1) x (-3) = -15+(n-1) x 2
=> 15 - 3n + 3 = -15 + 2n - 2
=> 12 - 3n = -17 + 2n
=> 12+17 = 2n+3n
=> 29=5n
=> n= 29/5
Correct Answer: 17280
Explanation:
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280
Correct Answer: 5040
Explanation:
The 7 letters word 'POVERTY' be arranged in ways = 7! = 5040 ways.
Correct Answer: 2520
Explanation:
Total number of letters in the word ABYSMAL are 7
Number of ways these 7 letters can be arranged are 7! ways
But the letter is repeated and this can be arranged in 2! ways
Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.
Correct Answer: 3600
Explanation:
Let the beds be numbered 1 to 7.
Case 1 : Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5´5!ways = 600 ways.
Case 2 : Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6.
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju?s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.
The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways
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