-
Posts
171 -
Joined
-
Last visited
Everything posted by Dhamnekar Win,odd
-
I did those computations for case(1) r=2, n=2 and case(2)r=2 and n=3. For case(1) The number of distinguishable distributions of 2 indistinguishable objects putting into 2 cells is 1 (equation or formula 1) and the numbers of distinguishable distributions of 2 distinguishable objects putting into 2 cells are 2(equation or formula 2) For case(2) The number of distinguishable distributions of 2 indistinguishable objects putting into 3 cells is -1 (equation or formula 1) and the number of distinguishable distributions of 2 distinguishable putting into 3 cells is 0(equation or formula 2) So, now how can we change the order of summation and use the binomial formula to express A(r, n+1) or in our case A(2,3) as the difference of two simple sums?
-
Assuming equation(2) to hold, how to express A(r-k, n) in equation (1) accordingly? How to change the order of summation and use the binomial formula to express A(r, n+1) as the difference of two simple sums as feller suggested? How to replace in the second sum v+1 by a new index of summation and use the following important property of binomial theorem \binom{x}{r-1} + \binom{x}{r} = \binom{x+1}{r}?
-
Elements of Combinatorial Analysis
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
Corrected equation (1) A(r, n+1)= \displaystyle\sum_{k=1}^{r} \binom{r}{k} A(r-k, n) Corrected equation (2) A(r, n) = \displaystyle\sum_{v=0}^{n} (-1)^v\binom{n}{v}(n-v)^r Then the author W. Feller says to replace in the second sum v + 1 by a new index of summation and use important property of binomial theorem which I wrote in my original question -
Elements of Combinatorial Analysis
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
Some more information : This problem refers to the classical occupancy problem (Boltzmann-Maxwell statistics): that is, r balls are distributed among n cells and each of the n^{r} possible distributions has probability n^{-r} -
-
Combinatorial Analysis (Investments)
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Analysis and Calculus
I have made mistakes in the computations of answers to question (a) and (b). Corrected answer:(a) \binom{n + r -1}{r}= \binom{12}{9}= 220 where n=9 and r = 4 Corrected answer:(b) 220 + \binom{15}{13} + \binom{14}{12} + 2 \times \binom{13}{11} =220 +105 +91 + 156 =572 Hence, author's answers are correct. -
-
Momentum is used to sense the amount of force applied to a moving object. With the help of Momentum, you can know the nature of the applied force on an object. Momentum is usually represented by the product of the mass and velocity of a moving object. But in this case of quantum mechanics, the equation of momentum will be different. How is this computed? How is this momentum computed in Quantum mechanics?
-
Question on structure of Atom
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
Answers: REMOVED BY MODERATOR -
An x-ray photoelectron spectroscopy experiment with an unidentified element, X, displays an emission spectrum with four distinct kinetic energies: 5.9 \times 10^{-17} J, 2.53 \times 10^{-18} J, 2.59 \times 10^{-20} J, 2.67 \times 10^{-20} J (Assume the incident light has sufficient energy to eject any electron in the atom.) (a)Name all of the possible ground state atoms that could yield this spectrum. (b)Calculate the binding energy of an electron in the 2p orbital of element X if the x-rays used for the spectroscopy experiment had an energy of 2.68 \times 10^{-16} J (c)Consider both the filled and unfilled orbitals of element X. Determine the number of: (1) total nodes in a 4d orbital (2)angular nodes in the 2p_y orbital (3)degenerate 5p orbitals How to answer all these questions?
-
Can I ask here statistics question? If yes, under which forum?
-
Rates of reactions N_2O_4 \rightarrow 2 NO_2
Dhamnekar Win,odd replied to Dhamnekar Win,odd's topic in Homework Help
Answer to d) 1)∆H = 17.4 kJ/mol 2)∆H = 8.7 kJ/mol 3)∆H= 4.35 kJ/mol Answer to e) Mechanism for this reaction is 1)N_2 + 2 O_2 \rightarrow 2 NO_2 \Delta H^\circ_f = 66.4 kJ/mol 2) N_2O_4 \rightarrow N_2 + 2 O_2 \Delta H^\circ_f = -9.16 kJ/mol