COMBINATION
Combination: Combination is nothing but selection of objects/things/numbers from a set of objects/things/numbers. Here the order is not important. Some formulae to remember for the combination.
The number of combinations of ‘n; distinct objects taken ‘r’ at a time (0< r < n) is given by
C (n, r) = n! / r! (n-r)!
Note: C (n, n) = 1 = C (n,0); C (n, r) = C (n, r-1); C (n, r-1) + C (n, r) = C (n+1, r).
1.Find the number of diagonals in a polygon having 15 sides.
Ans: (15 x 14) / (2) - 15 = 90 diagonals. {(n x n-1) ÷2}- n = n(n-3)/2
2.There are 12 persons in a party and if each, two of them shake hands with each other, how many handshakes happen in the party.?
Ans: Two persons hand shake is counted as one. Hence there are C (12 ,2) ways ; we get the result as 66 after simplification using the formula.
3.A committee of 3 persons is to be constituted from a group of 2 men and 3 women in how many ways can this be done if the committee to have 1 man and 2 women.?
Ans: 1 man out of 2 men be selected in C (2,1) ways; 2 women be selected in C (3,2) ways. Answer is 2 x 3 = 6 Ways.
4. 4 cards are to be chosen from a pack of 52 playing cards. If 2 cards of Red and 2 cards of Black are to be chosen, find the number of ways.
Ans: There are 26 Red cards and 26 Black cards. 2 red cards from 26 red cards C (26, 2) ,2 black cards from 26black cards C (26, 2). So
C (26, 2) x C (26, 2) = 2 x (26! / 22! .4!) = 2 x 23 x 25 x 26 = 29900 Answer. Formula used.
5.Find the number of ways in which a committee of 6 members can be formed from 8 men and 4 Women, to have at least 3 women in the committee.
Ans: Given 8 Men and 4 Women. The committee to have 6 members, with the condition to have at least 3 women. So, there are 2 possibilities.
I 3 women, 3 men, II. 4 women, 2 men
Required Answer: C (4,3) x C (8, 3) + C (4, 4) x C (8, 2)
On simplification we get 4 x 56 + 28 = 252. Note; (C (4,4) =1, C (4,3) = C (4,1)
6.Find the number of straight lines that can be formed by joining 20 points of which 4 points are collinear.
Ans: 4 points are collinear, it forms a straight line, we get a straight line by joining 2 points. Hence solution is C (20, 2) --- C (4,2) + 1 = 196 – 6 +1 = 185.
Note Since given 4 points are collinear to be subtracted as they form C4,2) lines in case of not collinearity.
7. In an examination a student has to answer 10 out of 13 questions, it is compulsory to choose at least 4 questions from the first 5 questions. Find the number of choices.
Ans: Possibilities are I: 4 + 6 = 10; II 5 + 5 = 10
Hence, we get C (5, 4) x C (8 ,6) + C (5, 5) x C (8, 5),
After simplification we get 5 x 28 + 1x 56 = 140 + 56 = 196.
8.Let Tn be the number of all possible triangles formed by joining vertices of an n –sided regular polygon. If Tn+1 -- Tn = 10, then, find the value of n.?
Ans: We use the formula C (n, r) + C (n, r-1) = C (n+1, r) -----I
Given Tn+1 – T n = 10 ie. C (n+1,3) –C (n, 3) = 10; using the formula I, we get
C (n ,2) = 10, by solving we get n= 5 (we get Q equation where factors are
(n-5) and (n+4) so the answer. Is n = 5
9. Find the number of ways in which one can select three distinct integers between 1 and 30 both inclusive, whose sum is even.
Answer: there are two possibilities 1. All are even, 2. Two odd and one even.
There are 15 odd integers 15 even integers. Hence the number of ways is,
C (15,3) + C (15, 2) x C (15 ,1); C (15,3) = 455, C (15, 2) = 105, C (15 ,1) ,
455 + 105x 15 = 455 + 1575 = 2030 is the ANSWER.
10.Find the number of ways distributing 10 distinct coins among 2 (two) people such that each will get at least one coin.
Answer: What are the possibilities let us see.
Select one coin and give it one person, the remaining to the second person. This can be done in C (n,1) ways; in the same way select two coins give them to one person, and the remaining to the second person. This can be done in n(n,2) ways. Proceeding in these ways, the number of ways is
C (n,1) + C (n,2) + C(n;3) + Cn.4) + …………. + C (n, n-1), ----I
We know C(n,0) + C (n,1) + C(n,2) + C(n,3) + ……………C (n, n-1) = 2n, (sum of binomial coefficients C0+ C1+C2+ --------+ Cn)
Now adding and subtracting C (n.0) and C(n,n) , in equation I,we get ,
The answer as 2 n - 2 since C (n.0) + C (n, n) = 1 + 1 = 2.
No comments:
Post a Comment