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Combination is one of the methods in mathematics, used for the selection of objects or items. The Combination is a method for choosing items or objects from a set of items or objects, where the selection order is irrelevant. For instance, consider a trio of numbers: A, B, and C. The number of ways that we can choose two numbers from each subset is thus defined by combination.When there are many group sets in a case, the possibility of the set of combinations increases, however when there are fewer group sets, it is feasible to count the number of combinations.
Read more: Set Operations
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Key terms: Permutation and combination, Combination, Permutation, Factorial, Set, Elements, Ratio
Combination Definition
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The combination in mathematics is defined as “The set of objects or items where the order of selection does not matter”.The combination is where the order does not matter. For choosing ‘k’ items out of ‘n’ elements, it is known that the permutation is always bigger than the number of combinations. Here is the formula which is used to find the combination.
Combination Formula
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Suppose a set "A" contains "n" distinct items, "p" distinct elements of which must be selected. The factorial ratio of a set with 'n' items determines how many 'p' combinations can be created from that set.
| nCp = (n)(n−1)(n−2)…….(n−p+1)/(p−1)(p−2)....(1) |
It can be written in the factorial terms
nCp = n! / (n - p)! P!
In the above-mentioned equation,
- It says that n>p is true and If n<p, then the value nCp becomes equal to zero.
- ‘C’ denotes the Combination of items.
- ‘N’ refers to the total number of items in the selection.
- ‘P’ denotes the number of items that must be chosen for the given combination.
Read more:
| Relevant Concepts | |
|---|---|
| Rational number | Poissson’s ratio |
| Factorial | Difference between permutation and combination |
Relation between Permutation and combination
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A special case of a permutation is a combination. It is a permutation in which the order in which the elements were selected is not taken into consideration. Therefore, it is always true that there are fewer combinations than permutations. The formula for a given set of permutations is
| P(n,r) = n!/(n-r)! |
The number of possible combinations in which 'r' items can be selected from 'n' items is as follows:
| nCr= n! ( n−r)!n!(n−r)!\{n!}{(n - r)!} r! → (2) |
With the equations (1) and (2) It is clear that the following correlations can exist between the permutations and combinations:
| nCq = nPq / q! |
In the above-mentioned equation,
- The combination of "r" elements out of "n" items is known as "nCr."
- The permutation of item 'r' from item 'n' is called nPr.
- The total number of options for selecting is 'n'.
- During the selection process, "r" represents the number of objects chosen.
Combination Properties
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The properties of combination are as follows:
- nC0 = nCn = 1
- nC1 = nCn-1 = n
- nCr + nCr-1 = n+1 Cr
- nCr = nCn-r
- nCr = nPr / r!
To fully understand the relationship between permutation and combination, look at the theorem.
Theorem: nPr = nCr.r!
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Since r items in each combination can be rearranged in r! different ways, we have r! permutations for each combination of nCr.
Proof: nPr = nCr.r!
= [n!/r!(n-r)!].r!
= n!/(n-r)!
As a result, the theorem is correct.
Things to Remember
- Selection of the project is through combination which is one the mathematical methods used.
- The combination method is applied when the selection of the objects or item is irrelevant.
- It is true that there are fewer combinations than permutations.
- The ‘c’ in the combination formula represents a combination of items.
- In the combination formula,the total number of the items are considered and the number of the items which are selected for the combination.
- Total number of the items are represented by ‘n’.
- The number of the items which are selected for the combination is represented by ‘p’.
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Sample Questions
Ques. Pick out 3 cricket balls from the set of 10 cricket balls. (1 marks)
Ans. 10.C10,3=10!/(7! x 3!)=10 x 9 x 8 /(3 x 2 x 1)
=120
Ques. A group of 3 football players P, Q, R. A team consisting of 2 players is to be formed. In how many ways can we do so? (3 marks)
Ans. In a combination problem, we know that the sequence of arrangement or selection is irrelevant..
Thus PQ= QR, QR = RQ, and PR=RS.
There are 3 options for choosing a team.
By using a combination formula, we have:
3C2 = 3!/2! (3-2)!
= (3.2.1)/(2.1.1) =3
Ques. Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} having 3 elements. (3 marks)
Ans. The set presented here has ten items. In whatever order, we must create subsets of three elements. The first subset we choose, 1,2,3, is identical to 3,2,1 if we choose it. So, in this case, we'll apply the combination formula.
As a result, 10C3 - 1 represents the number of subsets with 3 elements.
= 10!/(10-3)!3!
= 10.9.87!/7!.3!
= 10.9.8/3.2
= 120 ways.
Ques. Find the value 15C8 + 15C9 – 15C5 – 15C7(2 marks)
Ans. (15C8 + 15C9 ) – (15C5 + 15C7 ) [nCr + nCr-1 = n-1Cr]
= 15C9 – 15C7 = 0 [\(\because\) 15C9 = 15C7]
Ques. Find the value nCo + nC1 + nC2 + ………..+ nCn(1 mark)
Ans. 2n
Ques. Find the value of n. if the values are given 2nC3 : nC3 = 11:1 (4 marks)
Ans. \(\frac{\lfloor 2n}{\lfloor 3 \lfloor{2n-3}} \times \frac{\lfloor 3 \lfloor n-3}{\lfloor n} = \frac{11}{1}\) ………..(Given)
\(\implies \frac{2n(2n-1)(2n-2)\lfloor 2n-3}{\lfloor 2n-3} \times \frac{\lfloor n-3}{n(n-1)(n-2)\lfloor n-3} = \frac{11}{1}\)
\(\implies \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = \frac{11}{1}\)
\(\implies \frac{4(2n-1)(n-1)}{(n-1)(n-2)} = \frac{11}{1}\)
\(\implies \frac{4(2n-1)}{n-2} = \frac{11}{1}\)
⇒ 11n – 22 = 8n – 4 ⇒ 3n = 18
⇒ n = 6
Ques. Find the value 10C7 + 10C5 (2 marks)
Ans. nCr + nCr-1 = n-1Cr
= 10C7 + 10C5 = 10+1 C 7
= 11C7 = \(\frac{\lfloor 11}{\lfloor 11-7 \lfloor 7}\)
Ques. Prove nCr + nCr-1 = n-1Cr (5 marks)
Ans. L.H.S
Ques. A box containing 5 yellow caps and 6 blue caps determine the number of ways in which 2 yellow and 3 blue caps can be selected. (4 Marks)
Ans. No. of yellow caps =5
No. of blue caps = 6
No. of selecting yellow caps = 2…………….(Given)
No. of selecting blue caps = 3
Total no. of selection = 5C2 x 5C3
\(= \frac {\lfloor 5}{\lfloor 5-2 \lfloor 2} \times \frac{\lfloor 6}{\lfloor 6-3 \lfloor 3}\)
\(\frac{5 \times 4 \times 3!}{3! \times 2} \times \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2}\) = 200
Ques. Find the number of possible ways to select 5 cards from a 52-card deck that contains exactly 1 ace. (4 Marks)
Ans. There are four aces and 48 other cards in a deck of 52 cards. In this case, we must choose exactly one ace and four other cards.
The variety of options for selecting one ace from a deck of four aces = 4C1
The number of possible methods to select 4 cards from the remaining 48 cards = 48C4
There are 4 other cards 48C4 that can be chosen in ways that correspond to one manner of 4C1 choosing an ace. However, there are methods for selecting aces, thus the necessary number of methods can be used = 4C1 x 48C4 = \(\frac{4}{1} \times \frac{48 \times 47 \times 46 \times 45}{1 \times 2 \times 3 \times 4}\) = 778320
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