Home > Practice > Arithmetic Aptitude > Permutations and Combinations > Miscellaneous
11. There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.

Discuss
Solution:
The number of triangle can be formed by 10 points = 10C3
Similarly, the number of triangle can be formed by 4 points when no one is collinear = 4C3
In the question, given 4 points are collinear,Thus, required number of triangle can be formed,
= 10C3 - 4C3
= 120 - 4
= 116
12.  In a party every person shakes hands with every other person. If there are 105 hands shakes, find the number of person in the party.

Discuss
Solution:
Let n be the number of persons in the party
Number of hands shake = 105
Total number of hands shake is given by nC2
Now,
According to the question,
nC2=105or, n!2!×(n−2)!=105or, n×(n−1)2=105or, n2−n=210or, n2−n−210=0or, n=15,−14
But, we cannot take negative value of n
So, n = 15
i.e. number of persons in the party = 15
13. The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 without repetition.

Discuss
Solution:
One digit positive numbers = 5
Two digit positive numbers = 25
Three digit positive numbers = 100
4 digit positive numbers = 300
5 digit positive numbers = 600
Six digit positive numbers = 600
Total positive numbers,
= 5 + 25 + 100 + 300 + 600 + 600
= 1630
14. In the next World cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against each others once. Four top teams of this round will qualify for the semifinal round, where they play the best of three matches. The Minimum number of matches in the next World cup will be:

Discuss
Solution:
The number of matches in first round,
= 6C2 +6C2
Number of matches in next round,
= 6C2
Number of matches in semifinals,
= 4C2
Total number of matches,
= 6C2 + 6C2 + 6C2 + 4C2 + 2
= 53
15. There are 10 person among whom two are brother. The total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers , is equal to:

Discuss
Solution:
Total number of ways = 7! × 2!
16. If letters of the work KUBER are written in all possible orders and arranged as in a dictionary, then the rank of the word KUBER will be:

Discuss
Solution:
Total words starting with B = 4! = 24
Total words starting with E = 4! = 24
Total words starting with KB = 3! = 6
Total words starting with KE = 3! = 6
Total words starting with KR = 3! = 6
If Starting word will be KUBER, then rank of KUBER,
= 24 + 24 + 18 + 1
= 67
17. A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting four digit code with the proper combination of each of the 4 rings. Maximum how many codes can be formed to open the lock ?

Discuss
Solution:
There are 9 non-zero digits to arrange themselves at 4 different position. Each letter can be arrange at different position in 9 different ways.
So, required number of ways,
= 9 × 9 × 9 × 9
= 94
18. 10 students are to be seated in two rows equally for the Mock test in a room. There are two sets of papers, Code A and Code B. each of two rows can have only one set of paper but different that from other row. In how many ways these students can be arranged ?

Discuss
Solution:
5 students can be seated out of 10 students in 10C5 ways
Remaining 5 will be seated in,
= 5C5 ways
Students of each row can be arranged as,
= 5! × 5! ways
Two sets of paper can be arranged themselves in,
= 2! ways
Thus, Total arrangement,
= 10C5 × 5! × 5! × 2
= 7257600
19. How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?

Discuss
Solution:
Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways
20. Find the number of ways in which 8064 can be resolved as the product of two factors?

Discuss
Solution:
Total number of ways in which 8064 can be resolved as the product of two factors is 24 as below:
(1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152), (8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96)

Previous

Next