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Permutation & Combination

 

(A) Fundamental Principle of counting :

If an event can occur in `m' different ways, following which another event can occur in `n' different ways, then total number of ways of simultaneous occurrence of both events in definite order is = m × n (can be extended to any no. of events)

 

(B) What's Permutation & Combination ?

(1) Permutation :   * Arrangement of things taken some or all at a time.

                             * Order of occurrence of events is important.

(2) Combination :  * Collection or selection of things taken some or all at a time.

                             * Order of occurrence of events in not important

Note : All GOD made things in general are treated to be different and all man made things are to be spelled whether like or different

 

(C) Factorial :

(1) n ! = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE = product of 1st `n' natural numbers ⇒ n ! = 1 × 2 × .......... × n

(2) (n - 1)! = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEEPermutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(3) factorial of negative numbers is not defined.

 

(D) Useful Theorems :

T-1 : Number of permutations of `n' distinct things taken r' at a time Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

T-2 : Numbers of combinations/selections of `n' distinct things taken `r' at a time Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

 

Note :- Derived Identities :

(1) ncr + ncr - 1 = n + 1cr

(2) ncr = ncn - r

(3) if ncx = ncy ⇒ x = y or x + y = 0

(4) npr = r ! . ncr

(5) (2n)! = 2n.n! {1.3.5........(2n - 1)}

 

(E) Formation of Groups :

(1) Number of ways of dividing (m + n) different things in two groups having `m' and `n' things are : Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE;Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(i) If m = n, then number of groups = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(ii) If `2n' things are to be equally distributed among 2 persons then, no. of ways = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE × 2!

(2) IIIy by (m + n + p) different things can be divided into 3 unequal groups is Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(i) If all groups are equal then number of ways = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(ii) If `3n' things are to be equally distributed among 3 persons then, number of ways= Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

Note :- This can be extended to any number of groups.

 

(F) Permutation of alike objects :

Number of permutation of `n' things taken all at a time out of which

(1) `p' are similar and of one kind

(2) `q' are similar and of second kind

(3) and rest `r' are all different = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

 

Note :- Be careful if you encounter the following language used in problems

→ Number of other ways

→ Number of ways of rearranging

→ If as many more words as possible

 

(G) Circular permutation :

(1) Number of circular permutations of `n' different things taken `r' at a time = ncr (r - 1)!

(2) If clockwise & Anti-clockwise arrangements are considered as same then, ncr Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(3) Number of circular permutations of `n' things out of which `p' are alike and rest are different = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

 

(H) Total number of combinations :

(1) Number of ways of selecting at least one thing out of `n' different things is

= nc1 + nc2 +.............+ ncn = 2n - 1

(2) Number of ways of selecting at least one thing out of (p + q + r +..........) things in which p are alike of one kind, q of second kind & so on is = (p + 1) (q + 1) (r + 1) ........... -1

 

(I) Number of ways in which N can be resolved as a product of 2 divisors :

(1) N = pa . qb ...........p & q are prime

= Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

(2) Number of ways in which `N' can be resolved as a product of 2 divisors which are relatively prime = 2n-1 , where n → number of primes involved in prime factorization of N.

 

(J) Maximizing ncr :

Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

 

(K) Dearrangement :

Number of ways in which `n' letters can be placed in `n' directed envelopes so that no letter goes into its own envelope is = n!Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

 

(L) Distribution of alike objects :

(1) Number of ways of distributing `n' identical things to `p' persons where each person can receive one, none or more things is = n + p - 1Cp - 1

(2) Number of ways of distributing `n' identical things to `p' persons where each person should receive at least one object is = n - 1Cp - 1

 

(M) Grid Problem :

 Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

 

Number of ways of reaching B,

starting from point A are = Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE.

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FAQs on Permutations and Combinations, Chapter Notes, Class 11, Maths - JEE

1. What is the difference between permutation and combination?
Ans. Permutation and combination are two different concepts in mathematics. Permutation is the arrangement of objects in a particular order. The order of arrangement matters in permutation. For example, if we have three letters A, B, and C, then all possible arrangements of these letters are ABC, ACB, BAC, BCA, CAB, and CBA. Combination, on the other hand, is the selection of objects without considering their order. The order of selection does not matter in combination. For example, if we have three letters A, B, and C, then all possible combinations of these letters are AB, AC, and BC.
2. How do we calculate permutations and combinations?
Ans. To calculate permutations and combinations, we use the following formulas: Permutation: nPr = n!/(n-r)!, where n is the total number of objects and r is the number of objects selected. Combination: nCr = n!/r!(n-r)!, where n is the total number of objects and r is the number of objects selected.
3. What are the real-life applications of permutations and combinations?
Ans. Permutations and combinations have various real-life applications, such as: - In computer science, permutations and combinations are used to calculate the number of possible outcomes of a program or algorithm. - In probability theory, permutations and combinations are used to calculate the probability of an event occurring. - In genetics, permutations and combinations are used to calculate the probability of certain traits being inherited. - In finance, permutations and combinations are used to calculate the number of possible investment portfolios. - In sports, permutations and combinations are used to calculate the number of possible outcomes in a tournament or league.
4. What is the difference between a permutation and a combination with repetition?
Ans. The main difference between a permutation and a combination with repetition is that in a permutation with repetition, an object can be selected more than once, whereas in a combination with repetition, an object can be selected only once. For example, if we have three letters A, B, and C, and we want to select two letters with repetition, then the possible permutations are AA, AB, AC, BA, BB, BC, CA, CB, and CC. The possible combinations are AA, AB, AC, BB, BC, and CC.
5. What is the difference between a permutation and a combination with replacement?
Ans. The main difference between a permutation and a combination with replacement is that in a permutation with replacement, an object can be selected more than once and in a particular order, whereas in a combination with replacement, an object can be selected more than once but without considering their order. For example, if we have three letters A, B, and C, and we want to select two letters with replacement, then the possible permutations are AA, AB, AC, BA, BB, BC, CA, CB, and CC. The possible combinations are AA, AB, AC, BB, BC, and CC, but without considering their order.
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