Factorial notation
The continues product of first $n$ natural numbers is called "n factorial" or "factorial n" and is denoted by $n!$
Thus, $n!=1\times 2\times 3...\times (n-1)\times n$
Examples: $1!=1$
$2!=1\times 2=2$
$3!=1\times 2\times 3=6$
$4!=1\times 2\times 3\times 4=24$
$5!=1\times 2\times 3\times 4\times 5=120$
Note that,
$n!=n\times (n-1)...3\times 2\times 1$
= $n\times (n-1)!$ for $n>1$
= $n\times (n-1)\times (n-2)!$ for $n>2$
= $n\times (n-1)\times (n-2)\times (n-3)!$ for $n>3$ and so on.
Examples: $5!=5\times 4!$
$5!=5\times 4\times 3!$
$8!=8\times 7!$
$8!=8\times 7\times 6!$ and so on.
Zero Factorial
According to the above definition, $0!$ makes no sense.
However, we define $0!=1$.
n! is not defined for negative integers or fractions.
Permutations
Consider two examples,
ex 1: All arrangements which can be made from the letters a,b,c by taking 2 at a time are
ab, ba, bc, cb, ca, ac
Thus, the number of arrangements of 3 different objects taken 2 at a time is 6.
ex 2: All permutations which can be made from the letters C,A,T by taking all at a time are
CAT, CTA, ATC, ACT, TAC, TCA
Thus, the number of arrangements of 3 different objects taken all at a time is 6.
In the above two examples, each arrangement is different i.e. the order of writing the letters is important.
In ex 1, each arrangement is called a permutation of 3 different letters taken 2 at a time.
In ex 2, each arrangement is called a permutation of 3 different letters taken all at a time.
A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
Permutations when all the objects are distinct
The number of permutations of $n$ different objects taken $r$ at a time, repetition of objects not allowed, is given by
$^{n}P_{r}=P(n,r)=n(n-1)(n-2)...(n-r+1)$ $\large =\frac{n!}{(n-r)!}$
where, $0<r\leq n$
Derivation of the formula for $^{n}P_{r}$
$^{n}P_{r}=\frac{n!}{(n-r)!}$
The number of permutations of $n$ different objects taken $r$ at a time is same as the number of ways in which $r$ vacant places can be filled from $n$ different objects.
1st place can be filled in $n$ ways.
2nd place can be filled in $(n-1)$ ways.
3rd place can be filled in $(n-2)$ ways.
4th place can be filled in $(n-3)$ ways.
similarly,
$r$th place can be filled in $n-(r-1)$ ways i.e. $(n-r+1)$ ways.
By fundamental principle of counting,
$^{n}P_{r}=n(n-1)(n-2)(n-3)...(n-r+1)$
[Multiply numerator and denominator by $(n-r)(n-r-1)...3\times 2\times 1$]
$=\frac{n(n-1)(n-2)(n-3)...(n-r+1)(n-r)(n-r-1)...3\times 2\times 1}{(n-r)(n-r-1)...3\times 2\times 1}$
$^{n}P_{r}=\frac{n!}{(n-r)!}$
The continues product of first $n$ natural numbers is called "n factorial" or "factorial n" and is denoted by $n!$
Thus, $n!=1\times 2\times 3...\times (n-1)\times n$
Examples: $1!=1$
$2!=1\times 2=2$
$3!=1\times 2\times 3=6$
$4!=1\times 2\times 3\times 4=24$
$5!=1\times 2\times 3\times 4\times 5=120$
Note that,
$n!=n\times (n-1)...3\times 2\times 1$
= $n\times (n-1)!$ for $n>1$
= $n\times (n-1)\times (n-2)!$ for $n>2$
= $n\times (n-1)\times (n-2)\times (n-3)!$ for $n>3$ and so on.
Examples: $5!=5\times 4!$
$5!=5\times 4\times 3!$
$8!=8\times 7!$
$8!=8\times 7\times 6!$ and so on.
Zero Factorial
According to the above definition, $0!$ makes no sense.
However, we define $0!=1$.
n! is not defined for negative integers or fractions.
Permutations
Consider two examples,
ex 1: All arrangements which can be made from the letters a,b,c by taking 2 at a time are
ab, ba, bc, cb, ca, ac
Thus, the number of arrangements of 3 different objects taken 2 at a time is 6.
ex 2: All permutations which can be made from the letters C,A,T by taking all at a time are
CAT, CTA, ATC, ACT, TAC, TCA
Thus, the number of arrangements of 3 different objects taken all at a time is 6.
In the above two examples, each arrangement is different i.e. the order of writing the letters is important.
In ex 1, each arrangement is called a permutation of 3 different letters taken 2 at a time.
In ex 2, each arrangement is called a permutation of 3 different letters taken all at a time.
A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
Permutations when all the objects are distinct
The number of permutations of $n$ different objects taken $r$ at a time, repetition of objects not allowed, is given by
$^{n}P_{r}=P(n,r)=n(n-1)(n-2)...(n-r+1)$ $\large =\frac{n!}{(n-r)!}$
where, $0<r\leq n$
Derivation of the formula for $^{n}P_{r}$
$^{n}P_{r}=\frac{n!}{(n-r)!}$
The number of permutations of $n$ different objects taken $r$ at a time is same as the number of ways in which $r$ vacant places can be filled from $n$ different objects.
1st place can be filled in $n$ ways.
2nd place can be filled in $(n-1)$ ways.
3rd place can be filled in $(n-2)$ ways.
4th place can be filled in $(n-3)$ ways.
similarly,
$r$th place can be filled in $n-(r-1)$ ways i.e. $(n-r+1)$ ways.
By fundamental principle of counting,
$^{n}P_{r}=n(n-1)(n-2)(n-3)...(n-r+1)$
[Multiply numerator and denominator by $(n-r)(n-r-1)...3\times 2\times 1$]
$=\frac{n(n-1)(n-2)(n-3)...(n-r+1)(n-r)(n-r-1)...3\times 2\times 1}{(n-r)(n-r-1)...3\times 2\times 1}$
$^{n}P_{r}=\frac{n!}{(n-r)!}$
0 Comments