Permutation and combination

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Basic principle of counting

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बस्तुहरुको गणना गर्नको लागि निश्चित प्रकारको नियमको प्रयोग गरिन्छ जसलाई Counting Principle भनिन्छ । जस्तै

📙	📙	📙	📙	📙
📙	📙	📙	📙	📙
📙	📙	📙	📙	📙

यहाँ, किताबको सँख्या गणना गर्नको लगी 1,2,3,...15 वा 3*5=5*3=15 कुनै एउटा principle को प्रयोग गर्न सकिन्छ, तर जसमध्ये 3*5=5*3=15 गणना लाई efficient counting मानिन्छ।
यसै गरि बस्तुहरुको गणना गर्न प्रयोग गरिने Counting Principle मध्य तलका दुईवटा principle लाई Basic principle of counting मानिन्छ।

Basic principle of counting

1. Addition Principle
   If a task consists of a sequence of choices in which there are p selections for the first choice, OR q selections for the second choice, then the task of making either one first or one second choice can be done in
   p+q ways
   जस्तै, दुईवटा pen र तिनवटा marker बाट एउटा pen वा एउटा marker selection गर्नको लागि 2+3 तरिका

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2. Multiplication Principle
   If a task consists of a sequence of choices in which there are p selections for the first choice, AND q selections for the second choice, then the task of making one first and one second choice can be done in
   p*q ways
   जस्तै, दुईवटा pen र तिनवटा marker बाट एउटा pen र एउटा marker selection गर्नको लागि 3*2 तरिका

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Factorial

The notation n! represents the product of first n natural numbers, i.e.,
n!=n×(n-1)×(n-2)× . . . ×3×2×1
is We read this n! symbol as ‘n factorial’.
1 !=1
2 !=2 × 1
3 !=3 × 2 × 1
4 !=4 ×3 × 2 × 1
and so on.
We define
0 ! = 1

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Examples

1. How many three-digit even numbers can be formed?
   

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   We know that
   There are 10 digits
   0,1,2,3,4,5,6,7,8,9
   So, it can be used to form three digit number.

   Place value table

   सय (Hundred)	दश (Ten)	एक (Unit)
   nine possibility (1,2,3,4,5,6,7,8,9)	ten possibility (0,1,2,3,4,5,6,7,8,9)	five possibility (0,2,4,6,8)

   So, all together, there are
   9*10*5=450 three digit even numbers

2. How many three-digit number can be formed from 3,4,5 if repetition is allowed?
   

   Complete Solution 👉 Click Here

   We know that
   There are 3 digits
   3,4,5
   So, it can be used to form three digit number.

   Place value table

   सय (Hundred)	दश (ten)	एक (Unit)
   three possibility (3,4,5)	three possibility (3,4,5)	three possibility (3,4,5)

   So, all together, there are
   \(3*3*3=3^3=27\) three digit numbers

3. How many three-digit number can be formed from 3,4,5 if repetition is NOT allowed?
   

   Complete Solution 👉 Click Here

   We know that
   There are 3 digits
   3,4,5
   So, it can be used to form three digit number.

   Place value table

   सय (Hundred)	दश (ten)	एक (Unit)
   three possibility (3,4,5)	two possibility (3,4,5) one is used in 100 place	one possibility (3,4,5) one is used in 100 place one is used in 10 place

   So, all together, there are
   \(3*2*1=3!=6\) three digit umbers

4. How many three-digit number can be formed from 1,2,3,4,5 if repetition is allowed?
   

   Complete Solution 👉 Click Here

   We know that
   There are 5 digits
   1,2,3,4,5
   So, it can be used to form three digit number.

   Place value table

   सय (Hundred)	दश (ten)	एक (Unit)
   five possibility (1,2,3,4,5)	five possibility (1,2,3,4,5)	five possibility (1,2,3,4,5)

   So, all together, there are
   \(5 \times 5 \times 5=5^3=125\) three digit numbers

5. How many three-digit number can be formed from 1,2,3,4,5 if repetition is NOT allowed?
   

   Complete Solution 👉 Click Here

   We know that
   There are 5 digits
   1,2,3,4,5
   So, it can be used to form three digit number.

   Place value table

   सय (Hundred)	दश (ten)	एक (Unit)
   five possibility (1,2,3,4,5)	four possibility (1,2,3,4,5)	three possibility (1,2,3,4,5)

   So, all together, there are
   \(5 \times 4 \times 3=\frac{5!}{(5-3)!}=60\) three digit numbers

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Permutation

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Permutations is concerned with determining the number of different ways of arranging objects out of a given number of objects, without actually listing them. There are some basic counting techniques which will be useful in determining the number of different ways of arranging oobjects in definite order.
Therefore,
A permutation is a mathematical tool to arrange given objects in order. Permutations are for lists so order is important.

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Let us see some cases:

1. set of objects all different
2. repeated use of the same objects
3. set of objects not all different
4. circular arrangement

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Permutation: set of objects all different

1. The permutations of r objects taken all at a time out of n objects, repetition is NOT allowed is:
   \(\frac{n!}{(n-r)!}\)
2. The permutations of n objects taken all at a time out of n objects, repetition is NOT allowed is:
   \(n!\)

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Permutation: repeated use of the same objects

1. The permutations of r objects taken all at a time out of n objects, repetition is allowed is:
   \(r^n\)
2. The permutations of n objects taken all at a time out of n objects, repetition is allowed is:
   \(n^n\)

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Permutation: set of objects not all different (Permutation of repeated objects)

Assume that there are \(n_1\) objects of type 1, \(n_2\) objects of type 2,...,\(n_k\) objects of type k and \(n = n_1 + n_2 + · · · + n_k\) . The number of distinguishable permutations of these n objects is :
\(\frac{n!}{n_1! n_2!...n_k!} \)
This number is also the number of ways to place n distinct objects into k distinguished group with \(n_1\) objects in the first group, \(n_2\) in the second group,..., \(n_k\) in the last group.

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Permutation: circular arrangement

बस्तुहरुलाई circle मा arrange गर्नु परेमा, जस्तै n वटा बस्तुहरुलाई एउटा circle मा arrange गर्नु परेमा जम्मा (n-1)! वटा arrangement हरु बनाउन सकिन्छ।
The number of ways to arrange n distinct objects along a fixed circle is
\(\frac{n!}{n} = (n-1)! \)

Examples

1. Circular permutation for 1,2,3
   तिनवटा बस्तुहरु 1,2,3 लाई circle मा arrange गर्नु परेमा २ वटा मात्र distinct circular permutations हरु बनाउन सकिन्छ। जस्तै
   {1,3,2} and {1,2,3}
2. Circular permutation for 1,2,3,4.
   
   चारवटा बस्तुहरु 1,2,3,4 लाई circle मा arrange गर्नु परेमा ६ वटा मात्र distinct circular permutations हरु बनाउन सकिन्छ। जस्तै
   {1,2,3,4}
   {1,2,4,3}
   {1,3,2,4}
   {1,3,4,2}
   {1,4,2,3}
   {1,4,3,2}

यहाँ circle लाई rotate गर्न सकिने, र rotate गर्न गर्दा उस्तै हुने भएकोले cyclic permutations of objects हरु equivalent हुन्छन जसले गर्दा \(\frac{n!}{n} = (n-1)! \) को प्रयोग गर्नु पर्दछ।

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Examples

1. How many three-digit even numbers can be formed?
   
2. How many three-digit number can be formed from 3,4,5 if repetition is allowed?
3. How many three-digit number can be formed from 3,4,5 if repetition is NOT allowed?
4. How many three-digit number can be formed from 1,2,3,4,5 if repetition is allowed?
5. How many three-digit number can be formed from 1,2,3,4,5 if repetition is NOT allowed?
6. How many three-digit number can be formed if repetition is NOT allowed?
   Solution
   Permutations of 10 digits taken 3 at a time is
   P(10,3)
   These permutations will include 0's at 100’s place. For example, 012, 055, . . ., etc are such numbers which are actually 2-digit numbers so must be subtracted from P(10,3) to get the required number.
   Here
   We fix 0 at the 100’s place and rearrange the remaining 9 digits taking 2 at a time. This number is
   P(9,2)
   So
   The total three digit number is
   P(10,3)-P(9,2)
7. How many different 5-letter words can be formed from the word DEFINITION? Here,
   number of letters n= 10
   letter N repeats twice , so \(n_1=2\)
   letter I repeats thrice , so \(n_2=3\)
   Thus, the number of arrangements is
   \(\frac{n!}{n_1!n_2!} =\frac{10!}{2!3!}= 2520 \) ways

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Combination

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