Selection and combination is one of the most important and frequently tested topics in quantitative aptitude. It is asked in almost every competitive exam including CAT, SSC CGL, SSC CHSL, Bank PO, Bank Clerk, Railway RRB and CSAT. A strong understanding of selection and combination concept, formulas and tricks is essential for scoring well in these exams. This post covers nCr formula, selection from identical objects, total number of combinations, factor theory concept and more — all explained with clear formulas and solved examples.
Permutation & Combination → Selection & Combination
Introduction of Selection/Combination:-
Let us assume that Suresh has 5 friends A, B, C, D, E and he want to invite 2 of them for a party.
So the question is how many ways he can invite 2 friends out of 5 friends.
First Let us do it manually ⇒
A B C D E
Number of ways ⇒
(A, B) (A, C) (A, D) (A, E)
(B, C) (B, D) (B, E)
(C, D) (C, E)
(D,E)
= 10 ways
Since here we are to select 2 out of 5 values which we can do manually but if these numbers ↑ then it will become difficult to do it manually. So we will do it with the help of a formula.
⦿ Number of ways of selecting ‘r’ articles from ‘n’ distinct article is :-
Example:- Suresh has a party at his home. He has 12 friends & want to invite only 3 of them. In how many ways he can do this?
Solution:-
\(12{C_3} = \frac{{12!}}{{3! \times 9!}} = \frac{{12 \times 11 \times 10 \times 9}}{{3 \times 2}}\)
= 220 ways Answer
⦿ Number of ways of selecting ‘r’ articles from ‘n’ distinct articles if ‘k’ article is always included :-
Example:- In how many ways kapil can call 5 of his 12 friends if amit & ankit are always included?
Solution:-
¹²⁻²C₅₋₂
= ¹⁰C₃
= 120 Answer
⦿ The total number of combination of n different things taken one or more at a time = 2ⁿ – 1
Proof:-
Method (1):-
Total distinct things = n
for ‘1ˢᵗ’ thing there are 2 ways of selection: included or excluded
similarly, for ‘2ⁿᵈ’ thing there are 2 ways of selection: included or excluded
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Similarly, for ‘nᵗʰ’ thing there are 2 wayss of selection: included or excluded
Hence total number of ways of selection:
= 2×2×2×………….n times
= 2ⁿ
Here point to remember is that in these 2ⁿ ways, there is also 1 way in which none of the article will be included & we are to select one or more article
Hence number of ways = 2ⁿ – 1
Method (2):-
number of ways of selecting 1 thing out of n things = ⁿC₁
number of ways of selecting 2 thing out of n things = ⁿC₂
number of ways of selecting 3 thing out of n things = ⁿC₃
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number of ways of selecting n thing out of n things = ⁿCₙ
∴ total number of ways = ⁿC₁ + ⁿC₂ + ⁿC₃ + ………. + ⁿCₙ
& by property of binomial coefficient we know that value of aboveww expression is 2ⁿ – 1
Hence number of ways = 2ⁿ – 1
Example:- Amit, a teacher in a school, has 15 students in his class. He has to select 1 or more students for the laboratory work. then in how many ways he can do this?
Solution:- Here n = 15
⇒ 2¹⁵ – 1
= 32768 – 1
= 32767 ways Answer
Selection from identical object/article.
The total number of ways of selecting one or more things from ‘p’ identical things of one type, ‘q’ identical things of another type, ‘r’ identical things of another type and ‘n’ distinct things is :
Proof:
Number of ways of selecting 1 thing out of p identical thing = 1
Number of ways of selecting 2 thing out of p identical thing = 1
Number of ways of selecting 3 thing out of p identical thing = 1
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Number of ways of selecting p thing out of p identical thing = 1
Number of ways of selecting 0 thing out of p identical thing = 1
Hence number of ways of selecting 0 or more things out of p identical things = (p + 1)
similarly,
number of ways of selecting 0 or more things out of q identical things = (q + 1)
number of ways of selecting 0 or more things out of r identical things = (r + 1)
Also number of ways of selecting 0 or more out of n different things =
2×2×2×2×………..n times = 2ⁿ
Hence number of ways of selecting 1 or more out of given different things = (p + 1)(q + 1)(r + 1)2ⁿ
Here in above value there is also 1 way where none of the thing is selected.
Hence number of ways of selecting 1 or more of the given things :
Example:- There are 3 identical pens & 4 identical books & 5 identical pencils & 6 different highlighters. In how many ways one or more of these can be selected?
Solution:-
Here p = 3
q = 4
r=5
n = 6
∴ Number of ways = (3 + 1)(4 + 1)(5 + 1)2⁶ – 1
= 7679 Answer
Factor Theory Concept ⇒
Consider a composite number n
⇒ n = aᵖbᑫcʳ………..
where a, b, c are prime numbers & p, q, r are natural numbers.Then
Number of factors of n is given by:-
Proof:- A factor of above number has power of a as a⁰ or a¹ or a² or a³ or ……….. aᵖ
similarly, A factor of above number has power of b as b⁰ or b¹ or b² or b³ or ……….. bᑫ
A factor of above number has power of c as c⁰ or c¹ or c² or c³ or ……….. cʳ
Hence factors having a⁰ or any value of a can be (p + 1)
similarly, factors having b⁰ or any value of b can be (q + 1)
factors having c⁰ or nay value of c can be (r + 1)
& when we take any combination of these then total number of factors = (p + 1)(q + 1)(r + 1)……..
where 1 & the number itself is included.
Example:- Find the total number of factors of 2⁴3²5³7⁵.
Solution:-
∴ Total numbers of factors = 5 × 3 × 4 × 6
= 360 Answer
Example:- Find the number of odd factors of 2⁴3²5³7⁵
Solution:-
∴ Total number of odd factors = 1×3×4×6
= 72 Answer
Example:- Find the number of even factors of 2⁴3²5³7⁵
Solution:-
∴ Total number of even factors = 4×3×4×6
= 288 Answer
Example:- Find the number of factors of 2⁴3²5³7⁵ that are divisible by 10.
Solution:-
∴ Total number of factors divisible by 10 = 4×3×3×6
= 216 Answer
Example:- Find the number of factors of 2⁴3²5³7⁵ that are not divisible by 10.
Solution:- required answer = Total factors – factor divisible by 10
=360 – 216
=144 Answer
Example:- Find the number of factors of 2⁴3²5³7⁵ that are divisible by 100.
Solution:-
∴ Total number of ways = 3×3×2×6
= 108 Answer