in Combinatory recategorized by
192 views
3 votes
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?
in Combinatory recategorized by
192 views

2 Answers

7 votes
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.
edited by
0 votes
0 votes

I | II

3 | 5

4 | 4

5 | 3

Hence, 5C3*7c5 + 5c4*7c4 + 5C5 * 7C3 answer.

Answer:

Related questions