$15 \cdot 228$ students each choose at least one of English, Mathematics, and History. Among them, the number of students who choose both Mathematics and English but not History is equal to the number of students who choose only Mathematics. There are no students who choose only English or only History, 6 students choose both Mathematics and History but not English, and the number of students who choose both English and History is 5 times the number of students who choose all three subjects. If the number of students who choose all three subjects is a non-zero even number, then the number of students who choose both Mathematics and English but not History is
(A) 5.
(B) 6.
(C) 7.
(D) 8.
(E) 9.
(16th American High School Mathematics Examination, 1965)
［Solution］Let the number of people who choose both English and Math be $n_{1}$, and the number of people who choose all three subjects be $n_{2}$, where $n_{1} \geqslant 0, n_{2}>0$ and $n_{2}$ is even.
From the right diagram, we can see: $2 n_{1}+6 n_{2}+6=28$, that is,
$$
n_{1}+3 n_{2}=11 \text {. }
$$

The only solution that fits the problem is $n_{1}=5, n_{2}=2$. Therefore, the answer is $(A)$.