# Quick Calculation 14.6b

A First Course in String Theory

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Construct explicitly all the states with $\alpha' M^2=2$ and count them, verifying that there are indeed a total of 3200 states. You may find the counting formula in Problem 12.11 useful.

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Let $N(n, k) = {n + k - 1 \choose k - 1}$, the number of ways to put n indistinguishable balls into k boxes.

p.318 “For open bosonic strings $\alpha' M^2 = N^\perp - 1$, …”

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When $\alpha' M^2 = 2, N^\perp = 3$, the cases are:

1. three $a_1^\dagger$‘s:

$N(3,24) = 2600$

2. one $a_2^\dagger$ and one $a_1^\dagger$:

$24 \times 24 = 576$

3. one $a_3^\dagger$:

24

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Total number of possible states for $N^\perp = 3$ is 3200.

— Me@2018-04-20 02:45:35 PM

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