Chapter 9: Q26E (page 404)

We based the exact probabilities of equation (9-9) on the claim that the number of ways of addingN distinct nonnegative integer quantum numbers to give a total ofM is ${M + N - 1)! / (M! (N - 1)!]$ . Verify this claim (a) for the case $N = 2, M = 5$ and $(b)$ for the case.
$N = 5, M = 2$

Short Answer

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Step-by-step solution

Step by step solution

formula for number of ways

In this problem, our task is to verify the formula:

$number of ways = \frac{(M + N - 1)!}{M! (N - 1)!}$

by listing down the possible combinations for which distinct integers can be added to make for the following cases:

(a) $N = 2; M = 5$ .

(b) $N = 5; M = 2$ .

substitute value of N and M

(a)

Plugging $N = 2$ and $M = 5$ into the formula, we get:

$\begin{array}{rcl} number of ways & = & \frac{(5 + 2 - 1)!}{5! (2 - 1)!} \\ = & \frac{6!}{5!} \\ = & 6 \end{array}$

Listing down all 6 non-negative integer combinations that add up to 5 we thus have:

$N_{1} + N_{2}$	M
$0 + 5 5 + 0 1 + 4 4 + 1 2 + 3 3 + 2$	5 5 5 5 5 5

(b) Step 3: substitute value of N and M

Plugging $N = 5$ and $M = 2$ into the formula, we get:

$\begin{array}{rcl} number of ways & = & \frac{(2 + 5 - 1)!}{2! (5 - 1)!} \\ = & \frac{6!}{2! 4!} \\ = & 15 \end{array}$

Listing down all 15 non-negative integer combinations that add up to 2 we thus have:

$N_{1} + N_{2} + N_{3} + N_{4} + N_{5}$	M
$0 + 0 + 0 + 1 + 1 0 + 0 + 1 + 0 + 1 0 + 1 + 0 + 0 + 1 1 + 0 + 0 + 0 + 1 1 + 0 + 0 + 1 + 0 1 + 0 + 1 + 0 + 0 1 + 1 + 0 + 0 + 0 0 + 1 + 1 + 0 + 0 0 + 1 + 0 + 1 + 0 0 + 0 + 1 + 1 + 0 0 + 0 + 0 + 0 + 2 0 + 0 + 0 + 2 + 0 0 + 0 + 2 + 0 + 0 0 + 2 + 0 + 0 + 0 2 + 0 + 0 + 0 + 0$	$2 2 2 2 2 2 2 2 2 2 2 2 2 2 2$

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We based the exact probabilities of equation (9-9) on the claim that the number of ways of addingN distinct nonnegative integer quantum numbers to give a total ofM is ${M + N - 1)! / (M! (N - 1)!]$ . Verify this claim (a) for the case $N = 2, M = 5$ and $(b)$ for the case.
$N = 5, M = 2$

Short Answer

Step by step solution

formula for number of ways

substitute value of N and M

(b) Step 3: substitute value of N and M

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We based the exact probabilities of equation (9-9) on the claim that the number of ways of addingN distinct nonnegative integer quantum numbers to give a total ofM is{M+N-1)!/M!(N-1)!. Verify this claim (a) for the caseN=2,M=5and(b)for the case.N=5,M=2

Short Answer

Step by step solution

formula for number of ways

substitute value of N and M

(b) Step 3: substitute value of N and M

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Mechanics and Materials

Dynamics

Physical Quantities And Units

Circular Motion and Gravitation

Further Mechanics and Thermal Physics

Modern Physics

Study anywhere. Anytime. Across all devices.

We based the exact probabilities of equation (9-9) on the claim that the number of ways of addingN distinct nonnegative integer quantum numbers to give a total ofM is ${M + N - 1)! / (M! (N - 1)!]$ . Verify this claim (a) for the case $N = 2, M = 5$ and $(b)$ for the case.
$N = 5, M = 2$