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1. What is the difference between permutations and combinations in JEE?
Ans. Permutations and combinations are fundamental concepts in combinatorial mathematics. In JEE, permutations refer to the arrangement of objects in a particular order, while combinations refer to the selection of objects without considering their order. Permutations are denoted by nPr, where n is the total number of objects and r is the number of objects taken at a time. Combinations are denoted by nCr, where n is the total number of objects and r is the number of objects selected.
2. How do I calculate the number of permutations in JEE?
Ans. To calculate the number of permutations in JEE, you can use the formula nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects taken at a time. Here, "!" denotes the factorial function which means multiplying a number by all the positive integers less than it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.
3. What is the formula for combinations in JEE?
Ans. The formula for combinations in JEE is given by nCr = n! / (r! * (n-r)!), where n is the total number of objects and r is the number of objects selected. This formula takes into account the fact that combinations do not consider the order of the selected objects.
4. Can you give an example of a permutation problem in JEE?
Ans. Sure! Let's consider a problem where we need to arrange 3 different books (A, B, C) on a shelf. The number of ways to arrange these books is given by 3P3 = 3! = 3 x 2 x 1 = 6. Therefore, there are 6 different permutations possible for arranging these books on the shelf.
5. How do I solve combination problems in JEE involving multiple selections?
Ans. When solving combination problems in JEE involving multiple selections, you can use the formula nC(r1, r2, ...) = n! / (r1! * r2! * ...), where n is the total number of objects and r1, r2, ... represent the number of objects selected from each group. For example, if you have to select 2 objects from a group of 4 red balls and 3 objects from a group of 5 blue balls, the number of combinations would be 4C2 * 5C3.
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