Counting Principle
बस्तुहरुको गणना गर्नको लागि निश्चित प्रकारको नियमको प्रयोग गरिन्छ जसलाई Counting Principle भनिन्छ । जस्तै
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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 प्रमुख मानिन्छ।
- 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🖋️ 🖋️ 🖍️ 🖍️ 🖍️ - Multiplication Principle
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🖋️ 🖋️ 🖍️ 🖍️ 🖍️
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
Example 1
How many three-digit even numbers can be formed?
Complete Solution 👉 Click Here
Permutation
The study of 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 with a particular order. Permutations are for lists so order matters.
Permutation को अध्ययन बस्तुहरुको संख्या निर्धारणसँग गर्ने कुरा सँग सम्बन्धित छ।
दिएको वस्तुहरूको संख्याबाट निश्चित क्रमका आधारमा केही वा पुरै वस्तुहरुलाई arrange (वस्तुहरुको arrangement लाई सूचीबद्ध नगरीकन) गर्नको लागी Permutation को आवश्यक पर्दछ। यसमा केही आधारभूत गणनाको तरिकाहरु छन् जसले
वस्तुहरूलाई क्रमको आधारमा व्यवस्थित गर्न उपयोगी tool को काम गर्छ।
त्यसैले
Permutation भनेको निश्चित क्रमका आधारमा वस्तुहरुलाई arrange गर्ने एक विशेष mathematical tool हो जसले क्रमको आधारमा list तयार गर्छ ।
Example 2
How many three-digit number can be formed from 3,4,5 if repetition is allowed?
The number of permutations of n objects taken all at a time, repetition is allowed is:
\(n^n\)
Complete Solution 👉 Click Here
Example 3
How many three-digit number can be formed from 3,4,5 if repetition is NOT allowed?
The number of permutations of n objects taken all at a time, repetition is allowed is:
\(n!\)
Complete Solution 👉 Click Here
Example 4
How many three-digit number can be formed from 1,2,3,4,5 if repetition is allowed?
The number of permutations of r objects taken all at a time out of n objects, repetition is allowed is:
\(r^n\)
Complete Solution 👉 Click Here
Example 5
How many three-digit number can be formed from 1,2,3,4,5 if repetition is NOT allowed?
The number of permutations of r objects taken all at a time out of n objects, repetition is NOT allowed is:
\(\frac{n!}{(n-r)!}\)
Complete Solution 👉 Click Here
Example 6
How many three-digit number can be formed if repetition is NOT allowed?First, we count the permutations of 10 digits taken 3 at a time. This is
P(10,3).
But, 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 =P(10,3)-P(9,2)
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.
Example 7
How many different 5-letter words can be formed from the word DEFINITION? Here, the 10 letters without repetition, can be arranged in\(10!\) ways
Among the 10 letters in the word "DEFINITION", the letter N repeats twice (can be arranged in 2! ways), the letter I repeats thrice (can be arranged in 3! ways), thus, these repetitions should be removed from the permutations.
Therefore, the number of arrangements is
\( \frac{10!}{2!3!}= 2520 \) ways
Permutation of circular objects
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
\(\frac{n!}{n} = (n-1)! \)
For example, 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}
Example 8
Principle of 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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