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15 Nov 2009, 11:48
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h2polo
Find the number of selections that can be made taking 4 letters from the word "ENTRANCE".
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32
This problem is not the one you'll see on real GMAT. As combinatorics problems on GMAT are quite straightforward. But still it could be good for practice.
We have 8 letters from which 6 are unique.
Possible scenarios for 4 letter selection are:
A. All letters are different;
B. 2 N-s and other letters are different;
C. 2 E-s and other letters are different;
D. 2 N-s, 2 E-s.
Let's count them separately:
A. All letters are different, means that basically we are choosing 4 letters form 6 distinct letters: 6C4=15;
B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;
D. 2 N-s, 2 E-s: 2C2*2C2=1.
15+10+10+1=36
Answer: B.
Finding in the above word, the number of arrangements using the 4 letters.
This one should go relatively easy after we solved the previous. So we have:
A. 15 4 letter words with all distinct letters. # of arrangements of 4 letter word is 4!, as we have 15 such words, then = 15*4!=360;
B. 10 4 letter words with two N-s and two other distinct letters. The same here except # of arrangements would be not 4! but 4!/2! as we need factorial correction to get rid of the duplications=10*4!/2!=120;
C. The same as above: 10*4!/2!=120;
D. 2 N-s and 2 E-s, # of arrangement=4!/2!*2!=6.
Total=360+120+120+6=606.
Hope it's clear.
02 Jan 2010, 12:42
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8. How many positive integers less than 10,000 are such that the product of their digits is 210?
(A) 24
(B) 30
(C) 48
(D) 54
(E) 72
210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).
So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.
{1,6,5,7} # of combinations 4!=24
{2,3,5,7} # of combinations 4!=24
{6,5,7} # of combinations 3!=6
24+24+6=54.
Answer: D.
10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three:
3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}...
3 vertical
2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}
So the final answer would be; 9C3-8=84-8=76
Answer: B.
Hope it's clear.
(A) 24
(B) 30
(C) 48
(D) 54
(E) 72
210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).
So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.
{1,6,5,7} # of combinations 4!=24
{2,3,5,7} # of combinations 4!=24
{6,5,7} # of combinations 3!=6
24+24+6=54.
Answer: D.
10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three:
3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}...
3 vertical
2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}
So the final answer would be; 9C3-8=84-8=76
Answer: B.
Hope it's clear.
03 Jan 2010, 05:44
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Bunuel
The solution provided by you is well explained.. However for the text marked in red - is there a faster way for solving the eq for k...? Since time crunch is a big issue in GMAT exam? Please let me know your views...
Thanks,
JT
