20 points lie on a plane, out of which 7 points are collinear. Find the number of pentagons that can be drawn?

A :

15373

B :

15474

C :

15383

D :

12298

Time Spent: 1040 seconds

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A pentagon may be formed by joining 5 points lying on a plane. However, if we choose 3 or more points which are collinear, then the pentagon will not be formed.

Therefore, here we have 13 non-collinear points and 7 collinear points.

Number of pentagons that can be drawn = 13 C 5 + 13 C 4 7 C 1 + 13 C 3 7 C 2

= 1287 + 715 × 7 + 286 × 21 = 1287 + 5005 + 6006 = 12298

The correct answer is D.

how is it 13 C 4 7 C 1 + 13 C 3 7 C 2

pls can u explain why isit so?

In the said question,  there 20 points lying on a plane, out of which 7 points are collinear and we have to find out the number of pentagons that can be drawn.

Solution:

We know that a pentagon may be formed by joining 5 points lying on a plane. Out of these 5 points, at least 3 points be non-collinear or there can be maximum two collinear points while drawing a pentagon. 

Therefore, number of pentagons that can be drawn using 13 non-collinear points and 7 collinear points =

number of pentagons that can be drawn using all non-collinear points + number of pentagons that can be drawn using 1 collinear and 4 non-collinear points + number of pentagons that can be drawn using 2 collinear and 3 non-collinear points ...(1)

Now, number of pentagons that can be drawn using all non-collinear points i.e., using all 13 non-collinear points =  13 C 5

Similarly, number of pentagons that can be drawn using 1 collinear out of 7 given collinear point and 4 non-collinear out of 13 given non-collinear points =  13 C 4 7 C 1

And, number of pentagons that can be drawn using 2 collinear out of 7 given collinear point and 3 non-collinear out of 13 given non-collinear points =  13 C 3 7 C 2

 

On putting the above respective value in (1), we get

 

Number of pentagons that can be drawn using 13 non-collinear points and 7 collinear points 

= 13 C 5 + 13 C 4 7 C 1 + 13 C 3 7 C 2

= 1287 + 5005 + 6006 = 12298

 

 

Here, the terms  7 C 2 , 7 C 1 , 13 C 4 , etc. are representing the combinations techniques. Their use and solutions will be taught to you in 11th class. So, please tell us the test in 9th class from where you have find the same question so that after going through it, we can altered it according to your cognitive level.

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