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- Question
In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.
Options- A. 525
- B. 535
- C. 545
- D. 555
- Correct Answer
- 535
ExplanationThe number of points of intersection of 37 lines is . But 13 straight lines out of the given 37 straight lines pass through the same point A.
Therefore instead of getting points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting points, we get only one point B.
Hence the number of intersection points of the lines is = 535
- 1. When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?
Options- A. 620
- B. 671
- C. 625
- D. 567 Discuss
- 2. How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
Options- A. 4050
- B. 3600
- C. 1200
- D. 5040 Discuss
- 3. How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?
Options- A. 216
- B. 45360
- C. 1260
- D. 43200 Discuss
- 4. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?
Options- A. 1260
- B. 1400
- C. 1250
- D. 1600 Discuss
- 5. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Options- A. 135
- B. 63
- C. 125
- D. 64 Discuss
- 6. How many necklace of 12 beads each can be made from 18 beads of different colours?
Options- A. 18!
- B. 18! x 19!
- C. 18!(6 x 24)
- D. 18! x 30 Discuss
- 7. How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?
Options- A. 120
- B. 360
- C. 240
- D. 424 Discuss
- 8. The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and atleast 4 bowlers?
Options- A. 1024
- B. 1900
- C. 2000
- D. 1092 Discuss
- 9. In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?
Options- A. 36
- B. 25
- C. 24
- D. 72 Discuss
- 10. How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines?
Options- A. 215
- B. 315
- C. 415
- D. 115 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 671
Explanation:
When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.
The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.
Therefore, the number of outcomes in which at least one die will show 3 = 1296 ? 625 = 671
Correct Answer: 5040
Explanation:
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
=
= 5040.
Correct Answer: 43200
Explanation:
There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.
The number of ways in which 9 letters can be arranged = = 45360
There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in = 180 ways.
In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in = 12 ways.
The number of ways in which the four vowels always come together = 180 x 12 = 2160.
Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200
Correct Answer: 1260
Explanation:
A team of 6 members has to be selected from the 10 players. This can be done in or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260
Correct Answer: 63
Explanation:
Required number of ways =
Correct Answer: 18!(6 x 24)
Explanation:
Here clock-wise and anti-clockwise arrangements are same.
Hence total number of circular?permutations: =
Correct Answer: 360
Explanation:
There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.
Number of 7 digit numbers = = 420
But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.
= = 60
Hence the required number of 7 digits numbers = 420 - 60 = 360
Correct Answer: 1092
Explanation:
We are to choose 11 players including 1 wicket keeper and 4 bowlers or, 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in = 840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in =252
Total number of ways of selecting the team = 840 + 252 = 1092
Correct Answer: 25
Explanation:
The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2, 2, 1
b. 3, 1, 1
Case a. Number of ways of achieving the first option 2 - 2 - 1
Two toys out of the 5 can be selected in ways. Another 2 out of the remaining 3 can be selected in ways and the last toy can be selected in way.
However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2
Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways = 15 ways
Case b. Number of ways of achieving the second option 3 - 1 - 1
Three toys out of the 5 can be selected in ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3 - 1 - 1 option is = 10 = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.
Correct Answer: 315
Explanation:
Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
Therefore, the total number of parallelograms formed = 7C 2 x 6C 2 = 315
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