By J.N. Gurtu, A. Gurtu
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Example text
N - 2) ! = n . (n - 1) . (n - 2) . (n - 3) ! (3) [VI] Formula for the Value of npr. From equation (1), we have nPr = n(n - 1) (n - 2) (n - 3) ... (n - r + 1) [n(n - 1) . (n - 2)(n - 3) ... (n - r + 1)] [en - r) ... 3 . 2 . 1] [en - r) (n - r - 1) ... 3 . 2 . 1] [Multiplying above and below by (n - r) . (n - r - 1) . 3 . 2 . 1] _ n . (n - 1) . (n - 2) . (n - 3) ... 3 . 2 . 1 (n - r) . (n - r - 1) ... 3 . 2 . 1 np = •• r n! (n - r) ! n! (n - r) ! (4) [VII] Important Formulae for Finding the Number of Permutations (i) Number of permutations of n distinct objects taken r at a time = npr.
6) Independent events. , when A happens, 'B' mayor may not occur. e. B is independent of A. For example, if we throw a coin and a dice simultaneously, then the "result of coin" is independent of "result of dice". So 'R' and 'e5' are two independent events. , happening of A very much affects the prob. of happening of B. So, in case of mutually exclusive events, the events cannot be independent also. [V] Addition Theorem of Probability (1) If A and B are any two events of the same random experiment, then probability of happening of atleast one event out of these is given by P«A + B) =P(A (") B) =P(A) + P(B) - P(AB) (2) If the events are mutually exclusive then both cannot happen simultaneously and hence P(AB) = O.
1 -log x. 1 = - (1 + log x) x y 1:Jt.. =-:? (1 + log x) or dx 4 =x-x (1 + logx)2 -x-x. 1 dx x = x-x (1 + log x)2 _ x-x-1 Now ~ = 0 means that 1 +logx = 0 or log x = - 1 =log (lie), so x = (lie). x] dx 2 1 X= __ (1)-;-1 <0 lie e 28 PHYSICAL CHEMISTRY-I So, x = lIe is a point of maxima. Maximum value = elle. Problem 4. of x 3 Solution. Suppose y = x 3 - 3x2 - 9 ax2 - 9. t. x, ~=3x2-6x ... t. x, 42=6x-6 dx 2 Atx =0, d2 dx 2 ( ~) x=o =6xO-6=-6<0 .. y = x 3 - 3x 2 - 9 is maximum atx x =0 in equation (0.