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A elderly man, in order to practise manual dexterity, draws
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25 Jul 2006, 01:57
25 Jul 2006, 01:57
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A elderly man, in order to practise manual dexterity, draws 50 triangles a day by connecting three of the solid points on the grid made up of 16 squares shown in the attachment. He starts this activity on January 1 and is careful not to skip a day. On which day will he be forced to retrace one of his previously drawn triangles?
(A) February 12 (B) February 13 (C) February 14 (D) February 15 (E) None of these
(A) February 12 (B) February 13 (C) February 14 (D) February 15 (E) None of these
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Re: A elderly man, in order to practise manual dexterity, draws
[#permalink]
25 Jul 2006, 04:58
25 Jul 2006, 04:58
Think it is like that
Total number of selecting 3 poins out of 25 is 2300 triplets. Now remove 5C3*10 for the colinear triplets in vertical and horizontal lines of the grid. Remove 5C3*2 for the 2 diagonals 4C3*4 for the lines parallel to the diagonals, and finally 3C3 for the outer parallel lines to the diagonals. In total we have: 2300-( 100+20+16+4)=>2160 triangles. divide by 20 to find in how many days the different triangles will be depleted.
Total number of selecting 3 poins out of 25 is 2300 triplets. Now remove 5C3*10 for the colinear triplets in vertical and horizontal lines of the grid. Remove 5C3*2 for the 2 diagonals 4C3*4 for the lines parallel to the diagonals, and finally 3C3 for the outer parallel lines to the diagonals. In total we have: 2300-( 100+20+16+4)=>2160 triangles. divide by 20 to find in how many days the different triangles will be depleted.
Re: A elderly man, in order to practise manual dexterity, draws
[#permalink]
25 Jul 2006, 05:48
25 Jul 2006, 05:48
Answer E
As shown in the figure there are (5*5)25 distinct points in the figure.
To draw each triangle you need 3 distinct points.
He cannot retrace a line already drawn. It means that he cannot select a combination of points for a line which has already been selected.
Now , you as look at this problen as follows
How many ways you can select a combination of 3 points from 25 points ?
Using the formula for Combination
25!
--------- = 2300 , So he can draw 2300 triangles without retracing.
3!(25-3)!
Old man draws 50 triangles a day , so he will need 46 days.
January has 31 days , So he would finish drawing all possible triangles by Feb 15th. On the 16th he would have to retrace . Hence E
As shown in the figure there are (5*5)25 distinct points in the figure.
To draw each triangle you need 3 distinct points.
He cannot retrace a line already drawn. It means that he cannot select a combination of points for a line which has already been selected.
Now , you as look at this problen as follows
How many ways you can select a combination of 3 points from 25 points ?
Using the formula for Combination
25!
--------- = 2300 , So he can draw 2300 triangles without retracing.
3!(25-3)!
Old man draws 50 triangles a day , so he will need 46 days.
January has 31 days , So he would finish drawing all possible triangles by Feb 15th. On the 16th he would have to retrace . Hence E
