VIDEO LECTURES ON CSIR NET JRF MATHEMATICS-STATISTICS

CSIR NET JRF MATHEMATICS VIDEO LECTURES.In This videos i solved part a questions of CSIR NET MATHEMATICS. Watch and Subscribe the channel if you like them

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VIDEO LECTURE ON PERMUTATIONS AND COMBINATIONS- BASIC PRINCIPLE OF COUNTING

VIDEO LECTURE ON PERMUTATIONS AND COMBINATIONS- BASIC PRINCIPLE OF COUNTING

https://www.youtube.com/watch?v=tmcLbCcXT_w

In the video I  explained about permutations and combinations-a basic principle of counting.Usually permutations and combinations comes under combinatorial analysis which is highly used in solving problems of probability theory.The basic principle of counting is simple and everyone must know it before going to probability.This video is useful for competitive exam preparation like IIT JAM,CSIR NET JRF,SET,GATE,IBPS,BANK PO and many other. Please share the video with your friends  and subscribe my youtube channel.

CSIR NET MATHEMATICS JUNE 2015 QUESTION PAPER

CSIR NET MATHEMATICS JUNE 2015 QUESTION PAPER

download the csir net mathematics december 2014 question paper

CSIR NET MATHEMATICS DECEMBER 2014 QUESTION PAPER is now available at http://mathexamtips.blogspot.in .Soon we are going to provide the clear solutions of CSIR NET MATHEMATICS DECEMBER 2014 QUESTION PAPER

download the csir net mathematics june 2014 question paper

CSIR NET MATHEMATICS JUNE 2014 QUESTION PAPER is now available at http://mathexamtips.blogspot.in .Soon we are going to provide the clear solutions of CSIR NET MATHEMATICS JUNE 2014 QUESTION PAPER

100 SOLVED QUESTIONS FOR CSIR,GATE, SET AND OTHER COMPETITIVE EXAMS-PART 3

100 SOLVED QUESTIONS FOR CSIR,GATE, SET AND OTHER COMPETITIVE EXAMS-PART 3

Question 10: Let $H=\{e,(12)(34)\}$ and $K=\{e,(12)(34),(13)(24),(14)(23)\}$ be subgroup of $S_4$, where $e$ denotes the identify element of $S_4$. Then

1. $H$ and $K$ are normal subgroup of $S_4$
2. $H$ is normal in $K$ and $K$ are normal in $A_4$
3. $H$ is normal in $A_4$ but not in $S_4$
4. $H$ is normal in $S_4$ but $H$ is not

100 SOLVED QUESTIONS FOR CSIR,GATE, SET AND OTHER COMPETITIVE EXAMS-PART2

100 SOLVED QUESTIONS FOR CSIR,GATE, SET AND OTHER COMPETITIVE EXAMS-PART 2

Question 6: The number of elements in the set $\{m:1\leq m \leq 1000, m \text{ and } 1000 \text{ are relatively prime} \}$ is
$1.~100 \quad 2.~250 \quad 3.~300 \quad 4.~400$
Answer: The question asks for $\phi(1000)$.

And $\phi(n)$ can be calculated using following formula.
1. $\phi(p_1^{m_1}p_2^{m_2} \cdots p_r^{m_r}) = \phi(p_1^{m_1})\phi(p_2^{m_2}) \cdots \phi(p_r^{m_r})$.
2. $\phi(p^m) = \begin{cases}2^{m-1} &\text{ if } p= 2 \\ p^m - p^{m-1} &\text{ if } p \neq 2\end{cases}$
Or equivalently it can be calculated by single formula as follows
$$\phi(n) = n \prod_{p|n}\left(1-\frac{1}{p} \right)$$

$\phi(1000)=\phi(2^3)\cdot \phi(5^3) = 2^2(5^3-5^2)=400$.