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Tamil Nadu Board of Secondary EducationHSC Science Class 11

Revision: Combinatorics and Mathematical Induction Mathematics HSC Science Class 11 Tamil Nadu Board of Secondary Education

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Definitions [1]

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

Theorems and Laws [6]

Theorem: The number of permutations of n objects, where p objects are of the
same kind and rest are all different = `(n!)/(p!)`.
In fact, we have a more general theorem.

Theorem: The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ..., pk are of kth kind and the rest, if any, are of different kind is 
`(n!)/(p_1!p_2!...p_k!)`

Theorem: The number of permutations of n different objects taken r at a time, where repetition is allowed, is `n^r`. 
The number of 3-letter words which can be formed by the letters of the word
NUMBER = `"^6 P_3= (6!)/(3!)`= 4*5*6= 120. Here, in this case also, the repetition is not
allowed. If  the repetition is allowed,the required number of words would be `6^3` = 216. 
The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly
`"^12 P_2= (12!)/(10!)`= 11*12= 132.

Permutations when all the objects are distinct:

Theorem: The number of permutations of n different objects taken r at a time, where 0 < r ≤ n  and the objects do not repeat  is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by `"^n P_r`.
Proof: There will be as many permutations as there are ways of filling in r vacant the n objects. The first place can be filled in n ways; following which, the second place can be filled in (n – 1) ways, following which the third place can be filled in (n – 2) ways,..., the rth place can be filled in (n – (r – 1)) ways. Therefore, the number of ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . .   (n – (r – 1)) or n ( n – 1) (n – 2)
... (n – r + 1). 
This expression for `"^n P_r`  is cumbersome and we need a notation which will help to reduce the size of this expression. The symbol n! (read as factorial n or n factorial ) comes to our rescue. In the following text we will learn what actually n! means.

Theorem: `"^n P_r`= `"^n C_r` r!, 0 < r ≤ n.
Proof:  Corresponding to each combination of `"^nC_r`, we  have r ! permutations, because r objects in every combination can be rearranged in r ! ways. 
Hence, the total number of permutations of n different things taken r at a time is `"^nCr` × r!. On the other hand, it is P n r . Thus
`"^n P_r` =`"^n C_r` * r!, 0 < r ≤ n.

Theorem: `"^nC_r` + `"^nC_r-1`= `"^(n+1)C_r` 
Proof:  We have `"^nC_r` + `"^nC_r-1= (n!)/[r!(n-r)!] + (n!)/[(r-1)!(n-r+1)!]`


= `(n!)/ [r*(r-1)!(n-r)!] + (n!)/[(r-1)!(n-r+1)(n-r)!]`


 = `(n!)/[(r-1)!(n-r)!] [(1/r) + 1/(n-r+1)]`


= `(n!)/[(r-1)!(n-r)!] * (n-r+1+r)/[r(n-r+1)]`

= `(n+1)!/[r!(n+1-r)!`


= `"^(n+1) C_r`

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