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3 votes

In 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?

7 votes

The answer is -

(ways of selecting $3$ out of $5$ from I and $5$ out of $7$ from II) $+$

(ways of selecting $4$ out of $5$ from I and $4$ out of $7$ from II) $+$

(ways of selecting $5$ out of $5$ from I and $3$ out of $7$ from II)

$$

(5 \mathrm{C} 3) \times(7 \mathrm{C} 5)+(5 \mathrm{C} 4) \times(7 \mathrm{C} 4)+(5 \mathrm{C} 5) \times(7 \mathrm{C} 3)=420

$$

NOTE that the following answer/approach would be WRONG:

Find the mistake in doing it in this way:

(ways of selecting the minimum $3$ from each part I and II and $2$ out of the remaining $6$ from both parts)

$$

(5 \mathrm{C} 3) \times(7 \mathrm{C} 3) \times(6 \mathrm{C} 2)=5250.

$$

This Mistake is The Most Common Mistake that students make.

The mistake is that it's over-counting.

The second method would count selecting questions $1,2,3$ from part I, $1,2,3$ from part II, and then $4$ and $5$ from part I again, multiple times. Look if we rearrange how we choose:

$1,4,5$ from part I, $1,2,3$ from part II, and $2,3$ from part I also.

The second method would consider these two as two different combinations while clearly, they are the same.

(ways of selecting $3$ out of $5$ from I and $5$ out of $7$ from II) $+$

(ways of selecting $4$ out of $5$ from I and $4$ out of $7$ from II) $+$

(ways of selecting $5$ out of $5$ from I and $3$ out of $7$ from II)

$$

(5 \mathrm{C} 3) \times(7 \mathrm{C} 5)+(5 \mathrm{C} 4) \times(7 \mathrm{C} 4)+(5 \mathrm{C} 5) \times(7 \mathrm{C} 3)=420

$$

NOTE that the following answer/approach would be WRONG:

Find the mistake in doing it in this way:

(ways of selecting the minimum $3$ from each part I and II and $2$ out of the remaining $6$ from both parts)

$$

(5 \mathrm{C} 3) \times(7 \mathrm{C} 3) \times(6 \mathrm{C} 2)=5250.

$$

This Mistake is The Most Common Mistake that students make.

The mistake is that it's over-counting.

The second method would count selecting questions $1,2,3$ from part I, $1,2,3$ from part II, and then $4$ and $5$ from part I again, multiple times. Look if we rearrange how we choose:

$1,4,5$ from part I, $1,2,3$ from part II, and $2,3$ from part I also.

The second method would consider these two as two different combinations while clearly, they are the same.