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Basics of Permutation and Combination (in Hindi)
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Permutation and combination, How to count easily

Manoj Chauhan is teaching live on Unacademy Plus

U
Mam plz do part-2 also
sir, which book is better arihant or cengage
MI
sir aapne bola that ki repetition nahin honi chahiye lekin 7000 to 7999 men kaafi numbers ka repetition hoga jise aapko 1000 se minus karna padega to uska kya apne uske baare men kuch kaha nahi....
Ds
sir the question which you mentioned in introduction how can be sloved with the help of permutations.
Ap
sir is anything is left after this course of p&c plzzz sir tell
sir mera class 10th me 43% hai to kya iit me addmission ho jai ga agar 75% + aie ga to...
1. PERMUTATIONS AND COMBINATIONS INTRODUCTION Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with 10 digits from 0 to 9. The lock can be opened if 4 specific digits are arranged in a particular sequence with no repetition. Some how, you have forgotten this specific sequence of digits. You remember only the first digit which is 7. In order to open the lock, how many sequences of 3-digits you may have to check with?

2. To answer this question, you may, immediately., start listing all possible arrangements of remaining digits taken 3 at a time. But, this method will be tedious, because the number of possible sequences may be large. Here, in this Chapter, we shall learn some basic counting techniques which will enable us to answer this question without actually listing 3-digit arrangements. In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. As a first step, we shall examinea principle which is most fundamental to the learning of these techniques.

3. The factorial Factorial Notation Let n be a positive integer. Then, the continued product of first n natural numbers in called factorial n, to be denoted by n !, or n. Also, we define 0 ! 1. When n is negative or a fraction, n! Is not defined. Thus, n ! = n (n-1) (n-2) 3.2 . 1.

4. Q. 6.=(6 5 4 3 2 1) = 720, 3, = 3 2 x 1 = 6 and 1 ! = 1.

5. Solution =n(n-1)(n-2)(n-3) 3. 2 . 11 Thus, we have (i) n!- n (n -1) (n - 2)... 3 x 2x 1.

6. Q. Compute: 10! 30! 11!-10! (71)-(31) (6) HI 10

7. Solution We have 10!10x9x8x(7!) 10x9x8 (7)x(3!) (7!)x(3x2x1) 3x2xl = 120 30! 30x 29x (28!) (ii) 28! =-(28!) 30 29 = 870 119:10: 11-10-191.-10. 9:) . (110-10)(91. (9!) 100