The exact probabilities of equation (9-9) rest on the claim that the number of ways of addingNdistinct non-negative integer to give a total ofM is (M+N-1)![M!(N-1)!] . One way to prove it involves the following trick. It represents two ways that Ndistinct integers can add toM-9 and5, respectively. In this special case.

The X's represent the total of the integers, M-each row has 5. The 1'srepresent "dividers" between the distinct integers of which there will of course be N-Ieach row has8 . The first row says thatn1 is3 (three X'sbefore the divider between it andn2 ), n2is0 (noX's between its left divider withn1 and its right divider withn3 ),n3 ) is1 . n4throughn6 are0 , n7is1 , and n8and n9are0 . The second row says that n2is 2. n6is 1, n9is2 , and all othern are0 . Further rows could account for all possible ways that the integers can add toM . Argue that properly applied, the binomial coefficient (discussed in AppendixJ ) can be invoked to give the correct total number of ways for anyN andM .

Short Answer

Expert verified

The value forP=NM+Ne-nln(1+NM).

Step by step solution

01

Given data

The binomial coefficient can be utilized to calculate the number of ways for whichN nonnegative integers can add up toM .

02

concept of theK>>k  Approximation. 

K!(K-k)!Kk

03

Determine the K>>k Approximation. 

P=M!(N-1)!((M-n)+(N-1)-1)!(M-n)!(N-2)!(M+N-1)!=M!(M-n)!×(N-1)!(N-2)!×(M+N-1-(n+1))!(M+N-1)!=M!(M-n)!×(N-1)(N-2)!(N-2)!×(M+N-1-(n+1))!(M+N-1)!=(N-1)×M!(M-n)!×(M+N-1-(n+1))!(M+N-1)!

Solving further, we get

P=(N-1)×Mn×1(M+N-1)n+1=(N-1)Mn(M+N-1)n+1=NMn(M+N)n+1=NMn(M+N)(M+N)n

Simplifying further we get,

P=NM+N×Mn(M+N)n=NM+N×MM+Nn=NM+N×M+NM-n=NM+N×1+NMn=NM+N[eln(1+NM)]-n=NM+Ne-nln(1+NM)

Thus, the value of P is NM+Ne-nln(1+NM)

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Most popular questions from this chapter

Show that equation (9- 16) follows from (9-15) and (9- 10).

The Debye temperature of copper is 45K .

(a) Estimate its molar heat capacity at 100 K using the plot in Figure 9.33(b) .

(b) Determine its corresponding specific heat and compare it with the experimental value of 0.254J/g·K.

The diagram shows two systems that may exchange both thermal and mechanical energy via a movable, heat-conducting partition. Because both Eand Vmay change. We consider the entropy of each system to be a function of both:S(E,V). Considering the exchange of thermal energy only, we argued in Section 9.2 that was reasonable to define1TasδSδE. In the more general case, PTis also defined as something.

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b) Given this relationship, show thatdS=dQT(Remember the first law of thermodynamics.)

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(b) Stars several times our Sun's mass has sufficient gravitational potential energy to collapse further than a white dwarf; they can force essentially all their matter to become neutrons (formed when electrons and protons combine). When they cool off, an energy balance is reached like that in the white dwarf but with the neutrons filling the role of the incompressible fermions. The result is a neutron star. Repeat the process of Exercise 89. but assume a body consisting solely of neutrons. Show that the equilibrium radius is given by

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(c) Show that the radius of a neutron star whose mass is twice that of our Sun is only about10km .

From elementary electrostatics the total electrostatic potential energy in a sphere of uniform charge Q and radius R is given by

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