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- Question
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
- Correct Answer
- 35
ExplanationA triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points
The number of triangles formed = = 35
- 1. 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
- 2. 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
- 3. 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
- 4. If nC10=nC12 then,find n.
Options- A. 10
- B. 12
- C. 22
- D. 24 Discuss
- 5. Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.
Options- A. 28
- B. 14
- C. 21
- D. 7 Discuss
- 6. 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 Discuss
- 7. 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
- 8. 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
- 9. In how many different ways can the letters of the word 'POVERTY' be arranged ?
Options- A. 2520
- B. 5040
- C. 1260
- D. None Discuss
- 10. In how many different ways can the letters of the word 'ABYSMAL' be arranged ?
Options- A. 5040
- B. 3650
- C. 4150
- D. 2520 Discuss
Permutation and Combination problems
Search Results
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: 28
Explanation:
Let x, y, z be the number of balls received by the three persons, then
Let
x + y + z =21
u + 5 + v + 5 + w + 5 = 21
u + v + w = 6
Total number of solutions =
Correct Answer: 1170
Explanation:
Possibilities 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
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.
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