A bookshelf contains $6$ distinct books of Mathematics and $5$ distinct books of Physics. From these $11$ books, $6$ books are chosen at random. Let $X$ be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If $\alpha$ is the mean of the random variable $X$, then the value of $77\alpha$ is ________.

$\text{Maths: } 6 \text{ distinct books},\quad \text{Physics: } 5 \text{ distinct books}$ $6 \text{ books chosen at random from } 11$ $X=\bigl|(\text{\#Maths chosen})-(\text{\#Physics chosen})\bigr|$ $\text{To find: } \alpha=E[X],\ \text{and then } 77\alpha$ $1.\ m=\#\text{Maths chosen},\ p=\#\text{Physics},\ m+p=6$ $2.\ \text{ways}=\dbinom{6}{m}\dbinom{5}{6-m} \qquad 3.\ E[X]=\sum x\,P(x)$ $m=\#\text{Maths},\quad p=\#\text{Physics},\quad m+p=6$ $X=|m-p|=|m-(6-m)|=|2m-6|$ $p\le 5 \Rightarrow m\ge 1;\qquad m\le 6 \Rightarrow m\in\{1,2,3,4,5,6\}$ $X=|2m-6|:\ \ m=1\to4,\ 2\to2,\ 3\to0,\ 4\to2,\ 5\to4,\ 6\to6$ $\dbinom{11}{6}=462$ $w(m)=\dbinom{6}{m}\dbinom{5}{6-m}$ $m=1:6,\quad m=2:75,\quad m=3:200$ $m=4:150,\quad m=5:30,\quad m=6:1$ $6+75+200+150+30+1=462\ \ \checkmark$ $X=0:200,\qquad X=2:75+150=225$ $X=4:6+30=36,\qquad X=6:1$ $P(X=0)=\dfrac{200}{462},\qquad P(X=2)=\dfrac{225}{462}$ $P(X=4)=\dfrac{36}{462},\qquad P(X=6)=\dfrac{1}{462}$ $E[X]=\dfrac{0\cdot200+2\cdot225+4\cdot36+6\cdot1}{462}$ $E[X]=\dfrac{450+144+6}{462}=\dfrac{600}{462}$ $\alpha=E[X]=\dfrac{600}{462}=\dfrac{100}{77}$ $77\alpha=77\cdot\dfrac{100}{77}=100$ $\boxed{77\alpha=100}$