In an examination, a question paper consists of $12$ questions divided into two parts i.e., Part $\mathrm{I}$ and Part $\mathrm{II}$, 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?
In an examination, a question paper consists of $12$ questions divided into two parts i.e., Part $\mathrm{I}$ and Part $\mathrm{II}$, 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?
It is given that the question paper consists of $12$ questions divided into two parts - Part $I$ and Part $II$, containing $5$ and $7$ questions, respectively.
A student has to attempt $8$ questions, selecting at least $3$ from each part. This can be done as follows.
$(a)$ $3$ questions from part $I$ and $5$ questions from part $II$
$(b)$ $4$ questions from part $I$ and $4$ questions from part $II$
$(c)$ $5$ questions from part $I$ and $3$ questions from part $II$
$3$ questions from part $I$ and $5$ questions from part $II$ can be selected in $^{5} C _{3} \times^{7} C _{5}$ ways.
$4$ questions from part $I$ and $4$ questions from part $II$ can be selected in $^{5} C _{4} \times^{7} C _{4}$. Ways.
$5$ questions from part $I$ and $3 $ questions from part $II$ can be selected in $^{5} C_{5} \times^{7} C_{3}$ ways.
Thus, required number of ways of selecting questions
$=^{5} C_{3} \times^{7} C_{5}+^{5} C_{4} \times^{7} C_{4}+^{5} C_{5} \times^{7} C_{3}$
$=\frac{5 !}{2 ! 3 !} \times \frac{7 !}{2 ! 5 !}+\frac{5 !}{4 ! 1 !} \times \frac{7 !}{4 ! 3 !}+\frac{5 !}{5 ! 0 !} \times \frac{7 !}{3 ! 4 !}$
$=210+175+35=420$
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