Proving Permutation and Combinations Statements

Question 1

Question:   Prove that the number of ways in which (m+n) dissimilar things can be divided into two groups containing m and n

Answer:    If we select m things out of (m+n) things, then n things are left out . Then, this gives (m+n) that can be divided into two groups containing m and n things.

Question 2

Question:  

Answer:    i.


The value of C(n, 25) = C(25, 25) = 1

ii.



iii.




iv.



v.




\n = 8

Question 3

Question:  

Answer:   

= 5 + 10 + 10 + 5 + 1 = 31 = RHS

Question 4

Question:  

Answer:   



Aliter:

Question 5

Question:   A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Chemistry Part II, unless Part I is also borrowed. In how many ways can he choose the three books to be borrowed?

Answer:    There are two ways of selecting 3 books:

i) If Chemistry Part II is selected, then we have to select Chemistry Part I and in this case, the third book is selected from the remaining books.
Hence,the first selection can be made in C(6,1) ways = 6 ways.

ii) If Chemistry Part II is not selected, then we can select 3 books from the remaining 7 books. Hence second choice can be made in

Question 6

Question:   At an election, three wards of a town are canvassed by 4, 5 and 8 men respectively. If 20 men volunteer, in how many ways can they be allotted to the different wards?

Answer:    Let the three wards be labelled as A, B and C respectively. Let us select 4 men for ward A. 4 men can be selected out of 20 in ²⁰C₄ ways. Having selected 4 men from ward A in one of ²⁰C₄ ways, 5 men can be selected out of remaining 16 in ¹⁶C₅ ways for ward B. After the selection, we are left with 11 men. Now, 8 men have to be selected out of 11 in ¹¹C₈ ways for ward C.

Question 7

Question:   A candidate is required to answer 6 out of 10 questions, which are divided into two groups, each containing 5 questions, and he is not permitted to attempt more than 4 from each group. In how many ways can he make his choice?

Answer:   

Question 8

Question:   For the post of 5 teachers, there are 23 applicants. 2 posts are reserved for S.C. candidates and there are 7 S.C. candidates among applicants. In how many ways can the selection be made?

Answer:    There are 23 applicants. Among this, 7 are S.C. candidates. We have to select 2 S.C. candidates among 7 S.C. applicants.
So, there are 23 - 7 = 16 non-S.C. candidates.
We can select 2 S.C. canditates from 7 and the remaining 3 teachers from 16 applicants.
Hence, the number of ways in which the selections can be made is

Question 9

Question:   How many words of different 4 letters can be formed out of 7 capital letters, 3 vowels and 8 consonants, if each word starts with a capital letter and contains at least one vowel?

Answer:    Any capital letter can be chosen as the first letter of each word. So there are 7 ways of choosing the first letter of each word. Next, we are to choose 3 letters containing atleast 1 vowel as follows:

i. By choosing one vowel and two consonants


ii. By choosing 2 vowels and 1 consonant


iii. By choosing 3 vowels i.e.,³C₃ ways = 1 way.
\Total number of ways of choosing these 3 letters = 30+15+1 = 46
Further, the number of ways of arranging these 3 letters among themselves = 3! = 6 ways.

\The number of required words = (7)(6)(46) = 1932

Question 10

Question:   The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. Find the number of triangles that can be drawn using these points as vertices.

Answer:    The points 3+4+5=12 points lie in a plane.
To construct a triangle, we require three non-collinear points.
The number of selecting three points from these 12 points
=¹²C₃ = 220.
Number of selecting three points from three collinear points = ³C₃ = 1
(on AB)
Number of selecting three points from four collinear points = ⁴C₃ = 4
(on BC)
Number of selecting three points from five collinear points = ⁵C₃ = 10
(on AC)
Hence, the required number of = 220 - 1 - 4 - 10 = 205