Probability theory. 100 items and 3 controllers 100 items are checked by 3 controllers. What is probability that each of them will check more than 25 items? 
Here is full quotation of problem from workbook:
"Set of 100 articles randomly allocated to test between the three controllers. Find the probability that each controller has got to test at least 25 articles."
 A: Let $N_i$ denote the number of items checked by controller $i$. One asks for $1-p$ where $p$ is the probability that some controller got less than $k=25$ items. Since $(N_1,N_2,N_3)$ is exchangeable and since at most two controllers can get less than $k$ items, $p=3u-3v$ where $u=\mathrm P(N_1\lt k)$ and $v=\mathrm P(N_1\lt k,N_2\lt k)$. 
Furthermore, $v\leqslant uw$ with $w=\mathrm P(M\lt k)$ where $M$ is binomial $(m,\frac12)$ with $m=75$ hence $w\ll1$. And $N_1$ is binomial $(n,\frac13)$.
Numerically, $u\approx2.805\%$ and $w\approx0.1\%$ hence $1-p\approx1-3u\approx91.6\%$.
A: My try: $$\mathbb{P}(N_1\geq 25, N_2\geq 25, N_3\geq 25)=\frac{1}{3^{100}}\sum_{N_1\geq 25, N_2\geq 25, N_3\geq 25|N_1+N_2+N_3=100}\binom {100}{N_1,N_2,N_3}=$$
$$\frac{100!}{3^{100}}\sum_{N_1}\sum_{N_2}\sum_{N_3}\frac{1}{N_1!N_2!N_3!}=\frac{100!}{3^{100}}\sum_{N_1}\sum_{N_2}\frac{1}{N_1!N_2!(100-N_1-N_2)!}=$$
$$\frac{100!}{3^{100}}\sum_{N_1=25}^{50}\sum_{N_2=25}^{50-(N_1-25)}\frac{1}{N_1!N_2!(100-N_1-N_2)!}$$
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