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- Question
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
- Correct Answer
- 315
ExplanationParallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
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 - 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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 Discuss
- 6. The number of ways that 8 beads of different colours be strung as a necklace is
Options- A. 2520
- B. 2880
- C. 4320
- D. 5040 Discuss
- 7. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Options- A. 53400
- B. 17610
- C. 11760
- D. 45000 Discuss
- 8. A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most ?
Options- A. 215
- B. 268
- C. 254
- D. 216 Discuss
- 9. There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 6 choices each?
Options- A. 1112
- B. 2304
- C. 1224
- D. 2426 Discuss
- 10. In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?
Options- A. 4! x 4!
- B. 5! x 5!
- C. 4! x 5!
- D. 3! x 4! Discuss
Permutation and Combination problems
Search Results
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: 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: 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: 18!(6 x 24)
Explanation:
Here clock-wise and anti-clockwise arrangements are same.
Hence total number of circular?permutations: =
Correct Answer: 535
Explanation:
The 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
Correct Answer: 2520
Explanation:
The number of ways of arranging n beads in a necklace is = 2520
(since n = 8)
Correct Answer: 11760
Explanation:
Required number of ways = = = 11760
Correct Answer: 215
Explanation:
Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 - 1 = 215.
Correct Answer: 2304
Explanation:
Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6
Reuired total number of sequences
= 4 x 4 x 4 x 6 x 6
= 2304.
Correct Answer: 4! x 5!
Explanation:
The word EDUCATION is a 9 letter word, with none of the letters repeating.
The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants
As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.
The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.
Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.
Hence, the total number of ways = 4! × 5!
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