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

Torque and Rotational Motion

Thermodynamics

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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.
$N = 5, M = 2$