The post JEE Mathematics: Permutation and Combination appeared first on Magoosh JEE Blog.
]]>So, it is quite clear that to score decent marks in JEE Maths you should understand this topic completely and become comfortable in solving questions in this area. In this article, I will give you some basic theory, some tips, and some good examples which hopefully will help to strengthen your understanding.
If an operation can be performed in ‘m’ different ways and the second operation can be performed in ‘n’ different ways, then two operations in succession can be performed in ‘m * n’ ways.
To understand the above theorem clearly, consider this example:
The answer is obviously 15 (3*5). In this example, we use the Multiplication Theorem.
If two independent operations can be performed by m and n different ways respectively, then either of the operations can be performed in (m + n) ways.
To understand this theorem, consider rolling dice. There are two
independent events (operations): and for each operation we either get an odd face or even face. Both events can happen in 3 different ways, now you have to think in how many ways either of the two events can happen?The answer is 6 different ways. It is one of the simple use of Addition Theorem.
To understand the further concept in Permutation and Combination we have to know what a factorial is.
For any positive integer ‘n’, factorial of n can be defined as:
n!=n(n-1)(n-2)……1
Some examples- 4!=4*3*2*1=24
0!=1(standard)
I will discuss its definition later in this article.
A permutation is an arrangement of objects in a definite order.
Now, there are three different cases of permutations.
The number of permutations of n objects taken r at a time, where 0 < r ≤ n, denoted by ^{n}P_{r} , is given by
^{n}P_{r} = n! / (n-r)!
The number of permutations of n objects taken all at a time is n!.
For example, consider your class. Let’s assume that there are 50 seats and 50 students. So, the number of different ways in which all students can sit is 50!.
Explanation: The first student can sit in 50 different ways, the next one will sit in 49 different ways and so on, and all of them are dependent events so by using Multiplication Theorem total number of ways will be 50*49*48….*1=50!.
Another example is shuffling of cards. The number of ways in which total cards can be arranged is 52!.
The number of permutations of n objects, taken r at a time, when repetition of objects is allowed, is n^{r}.
Sometimes you may confuse what is n and r. So, you can remember it by thinking n as total and r correspond to want.
For example, think about a cycle lock which has a 4-digit code. Now we have to find how many possible lock combinations there are.
The answer is 10^4.
Explanation: First digit of lock can take any values from 0-9, so it can take 10 different values. Similarly, for 2nd,3rd, and 4th digit of the lock there will be 10 different values. Thus, by Multiplication Theorem, the total case will be 10*10*10*10=10^4.
The number of permutations of n objects of which p_{1} are of one kind, p_{2} are of the second kind, …, p_{k} are of kth kind and the rest if any, are of different kinds is
n! / (p_{1}!p_{2}!p_{3}!…p_{k}!)
Example- Most common examples of this topic are from arrangements of the letter.
Find the number of permutations by using the letters of ASSASSINS?
Explanation — ‘A’ repeats 2 times, ‘S’ repeats 5 times, and ‘I’ and ‘N’ occur a single time.
It is basically the number of ways in which one can select ‘r’ objects out of ‘n’ different objects where an order of selection is immaterial. It is denoted by ^{n}C_{r}.
^{n}C_{r} = n! / r!(n-r)!
The combination is a more important topic than permutation. Questions are generally asked from this topic.
As far as an example is concerned, you can think of many situations.
Consider the event of rolling a dice — you have to bet on any two numbers of the dice. The total number of combinations will be ^{6}C_{2}.
^{6}C_{2}=6! / 2!*4 =15.
You can also verify it by manually writing all the possible combinations. It will take very little time.
Sometimes a given question can cause confusion as to whether it is asking you to find permutations or combinations.
Some important results-
These results can easily be proved.
The above theories are the basics of permutation and combination. You can solve almost all questions of JEE Mains papers.
For JEE Advanced, you need to study a little bit more theory. They are basically the applications of what we have seen above. You can get all those concepts in any standard JEE preparation book.
I have discussed here mostly basic theories and examples. This is just a summary of what is present in the topic. So, you should read the NCERT one time and solve all its examples and problems. After that, you can refer to other books like NCERT Exemplar, and other JEE oriented books.
Permutation and Combination are a moderately hard topic. Initially, you may find it bit confusing, but once you start to get the idea it will be fun. Work hard, but enjoy the topic!
Best of luck!
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]]>The post JEE Mains Maths Syllabus: Complex Numbers and Quadratic Equations appeared first on Magoosh JEE Blog.
]]>The Complex Number came into existence because of our inability to solve some quadratic equations such as x^2 = -4. Complex Number is one of the feared topics by competitive exam aspirants, but it’s not that difficult from JEE point of view. All it requires is an indirect use of vectors and coordinate geometry. It’s best to practice more and more so that you are able to categorize different types of questions yourself. In most of the cases, you will have linked questions so all you need is a good understanding of the ‘concept’ rather than the tricks that can be applied in some specific situations only.
Quadratic Equations, on the other hand, being the most important and basic chapter in JEE Mathematics, is an easy topic of algebra. It is of importance largely because multiple questions in the other chapters are based on quadratic equations, which you cannot solve until and unless you have a good hold on this chapter. Quadratic equation inequalities form a major part of the problems in JEE.
Algebra part mainly focuses on the concepts of conjugate, modulus, argument and their properties. Sometimes question from this topic appear to be tricky, but by applying properties it can be reduced to the simplest form.
Geometrical interpretation covers rotation theorem (a bit tricky so may take some time), practice enough questions so that you know where and how to apply, it makes various solutions easy.
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