  Question

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

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