Theorem:
The number of circular permutations of n different objects is (n-1)..
Examples:
i) 1 + 4 + 7 + 10 + ... is a series in which first term is 1, second term is 4, third term is 7 and so on. ii) 3 - 9 + 27 - 81 + ... is also a series in which the first term is 3, second term is -9, third term is 27 and so ..
Note:
Strictly speaking 0! has no meaning. But since n P n = n! we may understand 0! = 1..
Fundamental Principle of Counting
Suppose two events E 1 and E 2 are to be performed in sequence, then if E 1 can be performed in 'm' ways and for each of these ways E 2 can be performed in 'n' ways, then the sequence E 1 E 2 can be performed in 'mn' different ways. This is known as the Fund..
Note:
In the above method note that To obtain D 1 , replace a 1 , a 2 , a 3 by d 1 , d 2 , d 3 in D To obtain D 2 , replace b 1 , b 2 , b 3 by d 1 , d 2 , d 3 in D To obtain D 3 , replace c 1 , c 2 , c 3 by d 1 , d 2 , d 3 in ..
In the above method note that To obtain D 1 , replace a 1 , a 2 , a 3 by d 1 , d 2 , d 3 in D To obtain D 2 , replace b 1 , b 2 , b 3 by d 1 , d 2 , d 3 in D To obtain D 3 , replace c 1 , c 2 , c 3 by d 1 , d 2 , d 3 in ..Co-factors
The co-factor of the element a ij is (-1) i+j times its minor a ij . We shall denote the cofactor of an element by the corresponding capital letter. Cofactor of a i j = A i j = (-1) i + j M i j Consider the determinant The minor of a 1 1 can b..
The co-factor of the element a ij is (-1) i+j times its minor a ij . We shall denote the cofactor of an element by the corresponding capital letter. Cofactor of a i j = A i j = (-1) i + j M i j Consider the determinant The minor of a 1 1 can b..Case II:
If D = 0 and D 1 , D 2 and D 3 are not all zero, then the system is inconsistent, that is the system has no solutio..
Summary
The fundamental principle of counting (F.P.C) states that if an operation can be performed in m different ways and if for each such choice, another operation can be performed in n different ways, then both operations, in succession can be performed in exactly mn different ways. The principle can ..
Example:
are all zero matrices of orders 1 x 2, 2 x 1, 2 x 2 and 3 x 3 respectivel..
are all zero matrices of orders 1 x 2, 2 x 1, 2 x 2 and 3 x 3 respectivel..Question 4
Question: Prove that n!(n + 2) = n! + (n + 1)!. Answer: L.H.S = n!(n + 2) = n![(n+1)+1] ..
Question: Prove that n!(n + 2) = n! + (n + 1)!. Answer: L.H.S = n!(n + 2) = n![(n+1)+1] ..See what our Users say :
Very fast and clear. Made sure I understood the concepts instead of giving the answers to the problem.
This Tutor Vista is GREAT! loved this session, it helped me heaps.
I like the way this tutors double check the work with me and took time to explain each detail. It really helped me to better understand - Harry
This is very helpful and explains everything thoroughly. when asked to do problems slowly, tutors did them very thoroughly and even went extra mile to explain how and why with text - I just bow down to Turor Vista
Looking for More Help!
