Given 5 history texts, 3 sociology texts, 6 anthropology texts and 4 psychology texts,

Question

Given 5 history texts, 3 sociology texts, 6 anthropology texts and 4 psychology texts, where all books are different, find the number of ways a student can choose

A) one of the texts,

B) two of the texts (hint: the order of the chosen books doesn't matter),

C) one history text and one other type of text,

D) one of each type of texts,

E) two different type of texts?

For A, we know that we have total $5+3+6+4$ texts, so it would be $18$.

For D, we know history (type) for $5$ different ways, sociology (type) for $3$ different ways, $6$ for anthropology and $4$ for psychology. Then it would be $360 = 5 \cdot 3 \cdot 6 \cdot 4$.

Can someone help me understand others while the solution is given as

B)two of the texts (hint: the order of the chosen books doesn't matter) $= 153$

C)one history text and one other type of text $= 65$

E)two different type of texts $= 119$.

Answer 1

• B): $18$ ways for the first, $17$ ways for the second, as order does not matter divide by $2!$ (or what is the same: $\binom{18}{2}$): $\boxed{\frac{18\cdot 17}{2} = 153}$
• C): $5$ ways for the history book, $18-5 = 13$ for the other book: $\boxed{5\cdot 13 = 65}$
• E): Subtract from B) all ways of getting $2$ of each type: $\boxed{\binom{18}{2} - \left( \binom{5}{2} + \binom{3}{2} + \binom{6}{2} + \binom{4}{2}\right) = 119}$