বিন্যাস কিভাবে খুজা যায়

The product of all consecutive integers from $1$ to a non-negative integer $n$ is known as $n$ factorial, denoted $(n!)$.

This means that for any non-negative integer $n$, $$n!=(1)\times(2)\times(3)\times ...\times(n-2)\times(n-1)\times(n)$$

Try this problem:

$4!=(1)\times(2)\times(3)\times(4)=24$

Here's another one:

As you can see, factorials get big pretty quickly!

Factorials are very useful in permutations and combinations.

A permutation is an arrangement of elements. For example, $TAMH$ is a permutation of $MATH$ because it is an arrangement of the elements $M, A, T,$ and $H$. In permutations, the order of the elements does not matter. So if $M$ comes before $A$, it will be a different permutation than if $A$ comes before $M$.

Try this problem!

Since there are $5$ choices for 1st place, $4$ choices for 2nd place, $3$ choices for 3rd place, $2$ choices for 4th place, and $1$ choice for 5th place, we find that there are $5\times4\times3\times2\times1=(5!)=120$ possible different finishing orders.

For a permutation where we are choosing $r$ things from $n$ total things, we find that the number of permutations is: $\frac{n!}{(n-r)!}$

Try this problem using the formula.

Just plug in $5$ for $r$ and $8$ for $n$ to get $\frac{8!}{(8-5)!}$ which simplifies to $8\times7\times6\times5\times4=6720$.

A combination is similar to a permutation except in combinations, order does matter. $MATH$ would be considered the same combination as $TAMH$ because you are not changing any of the elements, you are simply changing the order of the same elements.

A combination where you choose $r$ things from $n$ total things can be denoted with this formula: $$\frac{n!}{r!(n-r)!}$$

Okay now that you know a lot about factorials and combinatorics, here's a tricky question for you.

$(0!)=1$ because, in a combinatorial sense, we have exactly one way to permute zero objects.

What is a Permutation?

A permutation of a set of distinct symbols is an arrangement of the symbols in a line in some order.

Finding Permutations

Given natural numbers $k$ and $n$ with $k \leq n$, the number of ways to place $k$ marbles of distinct colors into $n$ numbered boxes, at most one marble per box, is $\frac{n!}{(n-k)!}$.

Or in other words, the number of possible permutations on $k$ objects from a set of $n$ is $\frac{n!}{(n-k)!}$.

The number of permutations of $n$ symbols is $n!$.

Example 1:

How many couples can be selected from a group of $12$ women and $20$ men?

The number of ways to pair up women (marbles) and men (boxes) such that no woman is paired with the same man more than once is $\frac{20!}{(20-12)!} = \frac{20!}{8!}$.

Example 2:

There are $7! = 5040$ ways in which $7$ people can form a line. How many ways can you arrange those $7$ people in a circle?

Your first though was probably the number of permutations on $7$ people is $7!$ however, that is wrong. If you are standing in a circle and every person shifts left or right they would be standing in different places in the circle but standing next to the same people. Thus, the order of people has not changed. This can happen $7$ times so the number of ways to arrange $7$ people line a circle is $$\frac{7!}{7} = \frac{7\cdot 6!}{7} = 6!.$$

Test Yourself!

Factorials!!!

A factorial is the multiplication of all of the numbers from $n$ to $1$ for any integer $n$. So basically $n! = n \cdot (n-1) \cdot (n-2) \cdot \cdots \cdot 2 \cdot 1$. So a number like $5!$ will be $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$. Factorials are quite simple!

Now try this one!

How many ways are there to order $6$ people in a line?

For simplicity, lets call these $6$ people: $A$, $B$, $C$, $D$, $E$ and $F$ respectively.

How many different people can be at the front of the line? Well, there are $6$ people and there are no constraints in this question, so there are $6$ different people that can be at the front of this line.

Lets say we decided who is going to be at the front of the line, how many different people can be the second in the line? Well, say that $A$ is at the front of the line, surely $A$ cannot be second in the line as well! Hence, $A$ cannot be second. The rest of the people are all possible candidates to be second. So there are $5$ different people that can be second in the line. This holds true no matter who is at the front of the line, not just $A$.

How about the third in line? If we already know who is going to be first and second, how many different people can be the third in line? Well, using the same argument as before, only $4$ people can be third in line.

Using the same argument yet again, there are $3$ people that can be forth in line; $2$ people that can be fifth in line; and only $1$ person can be last.

Hence, by the rule of product, the number of ways to order the $6$ people is: $6\times5\times4\times3\times2\times1=\boxed{720}$

For any positive integer $n$, $n!$ (pronounced $n$ factorial) is defined as:

$$n!=n\times(n-1)\times(n-2)\times(n-3)\cdots4\times3\times2\times1$$

The factorial of $n$ can be used to find the number of orderings for $n$ objects. However, this only works when the objects are distinct.

COMBINATIONS, PERMUTATIONS, AND FACTORIALS

Combinations and permutations are used in math to show the number of different ways a set of elements (numbers, objects, things, etc.) can be arranged.

Permutations are used when the order of the elements matter such as a code for a lock or final placements in a race. Combinations are used when the order of the group doesn’t matter, such as finding all the different group of toppings you could put on a pizza.

FACTORIALS

Before we get into calculating combinations and permutations, you must first understand what a factorial is. The dictionary definition of a factorial is “the product of an integer and all the integers below it.” Written out in general terms, the equation would be: $$n!=\left(1\right)\cdot\left(2\right)\cdot\left(3\right)\cdot...\cdot\left(n-2\right)\cdot\left(n-1\right)\cdot\left(n\right)$$ When you are finding a factorial, it is rather simple to calculate it, regardless of the looks of the above equation: $$\begin{align}3!&=3\cdot2\cdot1=6 \\ 4! &=4\cdot3\cdot2\cdot1=24 \\ 7! &=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=5040\end{align}$$ The only time this doesn’t hold true is with $0!$. If you ever come across $0!$ in a problem all you need to know is that $0! = 1$.

(Video) Permutations and Factorials (Simplifying Math)

By Eric Buffington

In this video, we look at a couple examples of permutations.

$\large{\text{Summary}}$

Factorials

Written as $6!$ $6!=6\times5\times4\times3\times2\times1=720$ The video then goes through some more examples of factorials.

Note: $0!=1$

Permutations

Arrangements when order matters.

Example: You have $3$ marbles. How many different patterns can you make?

For the first spot, you can put any of your $3$ marbles in that spot. After the first marble is placed you only have $2$ marbles to pick from. Finally, you are left with only $1$ option for the third spot. So, you have $3\times{2}\times{1}=6=3!$ options.

There is also a formula: $_nP_r=\frac{n!}{(n-r)!}$ where $P$ is the number of permutations, $n$ is the items taken, and $r$ is how many items are taken at a time.

There are some examples in the video using the formula. Try them out!