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An examination, a question paper consists of $$12$$ questions divided into two parts i.e., Part $$I$$ and Part $$II$$, containing $$5$$ and $$7$$ questions, respectively. A student is required to attempt $$8$$ questions in all, selecting at least $$3$$ from each part. In how many ways can a student select the questions?


Solution

There are $$5$$ question in parts & $$7$$ question in parts $$2$$
atleast $$8$$ question have to we attempted & $$3$$ from lach part.
There are $$3$$ options
$$(1)\quad 3$$ from part $$1$$ & $$%$$ from part $$2$$
no, of ways
$$=^5C_3 \times ^7C_5$$
$$=\dfrac {5!}{3! (5-3)!}\times \dfrac {7!}{5!(7-5)!}$$
$$=\dfrac {5\times 4}{2}\times =\dfrac {7\times 6}{2}$$
$$=5\times 2\times 7\times 3$$
$$=210$$ ways 

$$(2)\quad 4$$ from part $$1$$ & $$4$$ from part $$2$$
$$=^5C_4\times ^7C_4$$
$$=\dfrac {5!}{4! (5-4)!}\times =\dfrac {7!}{4!(7-4)!}$$
$$=5\times =\dfrac {7\times 6\times 5}{3\times 2\times 1}$$
$$=5\times 7\times 5$$
$$=175$$ ways

$$(3)\quad 5$$ from part $$1$$ & $$3$$ from part $$2$$
$$=^5C_5 \times ^7C_3$$
$$=\dfrac {5!}{5!\ 0!}\times =\dfrac {7!}{3! (7-3)!}$$
$$=1\times =\dfrac {7\times 6\times 5}{3\times 2}$$
$$=35$$ ways
Total number of ways $$=210+175+35$$
$$420$$ ways

Mathematics

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