How many different selections of $5$ books can be made from $12$ different books if,
- two particular books are always selected?
- two particular books are never selected?
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When $2$ particular books are always selected, then we have to select $3$ books from $10$ books.
$2$ particular books can always be selected from $12$ books in $1$ way
$3$ books can be selected from $10$ books in
${^{10}}{C}_{3} = \dfrac{10!}{7! \times 3!} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ ways
$\therefore \;$ Number of ways of selecting $5$ different books from $12$ books when two particular books are always selected $= 1 \times 120 = 120$ ways
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When two particular books are never selected, then we have to select $5$ books from $10$ different books.
$5$ books can be selected from $10$ books in
${^{10}}{C}_{5} = \dfrac{10!}{5! \times 5!} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ ways