[insane]

hey newkid, cool to see ur score... congrats...
through some light on where did u start ur preperation from? I had prepared for CAT, IIMs, had some calls but couldnot make it. Now I am planning to take up GMAT. gimme a starting thread...

Originally Posted by newkid

congratulations on a great score.

i am bit confused about your scores though
i myself appeared for gmat on 25.04.05 and my score was as under

Q - 50
V - 41

Overall - 760

Hence, i c that though my quant scaled score is same as yours, my verbal scaled score is one less than yours, still my overall score is more than u. INTERSTING, isn't it?

[sumit82]

Hi All,

Somebody would have done the question..........

Q. How many shortest paths are there from City A to City B
Options
1. 8
2.10
3.14
4.20

Clue to rem..........The shortest path length is 5 ... so it consists of 5 movements.... 2 right
... 3 up

The way shown in ETS or rather OG is to list all possible ways of arranging U and R

But I think this question can be done using permutation & combination.

Can someone throw light on that alternative plz...


Cheers
Sumit

[newkid]

answer can be obtained using permutations

it is 5!
3! 2!

total no. of steps is 5
3 r of one kind and 2 are of other kind

hence the result.

[sumit82]

Originally Posted by newkid

answer can be obtained using permutations

it is 5!
3! 2!

total no. of steps is 5
3 r of one kind and 2 are of other kind

hence the result.

Hi

1. Can Combinations be used to arrive at this answer.....

2. What if I try and place them

_ _ _ _ _

3c1 x 2c1 X 1c1 X 2c1 X 1c1 = 12

But then I do have to arrange them too in 5! ways, so derived ans is wrong

Cheers

[newkid]

It is a permutation problem and analgous to rearrangement of letters of a word. for example. how many differnt words can be made by rearranging the letters of the word PEPOLE
answer to the said problem is not 6! but 6! divided by 2! and again by 2!
i.e. 6!
2!2!

it is so because of the repitions of the letters P (2 REPITITIONS) AND E (AGAING 2 REPITIONS) . SIMILARLY, in the given problem , the no. of steps is 5 with 3 in one direction and 2 in other direction with the restriction of using a particular step more than once. Hence, the given problem translates in a type as highlighted above with 5 steps and reptiotions of 3 steps and 2 steps.


i think now, u get the solution

p.s. - by the way anwer to the problem is 10 and not 12. u can obtain by solving as stated.

[sumit82]

Originally Posted by newkid

It is a permutation problem and analgous to rearrangement of letters of a word. for example. how many differnt words can be made by rearranging the letters of the word PEPOLE
answer to the said problem is not 6! but 6! divided by 2! and again by 2!
i.e. 6!
2!2!

it is so because of the repitions of the letters P (2 REPITITIONS) AND E (AGAING 2 REPITIONS) . SIMILARLY, in the given problem , the no. of steps is 5 with 3 in one direction and 2 in other direction with the restriction of using a particular step more than once. Hence, the given problem translates in a type as highlighted above with 5 steps and reptiotions of 3 steps and 2 steps.


i think now, u get the solution

p.s. - by the way anwer to the problem is 10 and not 12. u can obtain by solving as stated.

Hi newkid,

I understand the answer is 10 and the sol u have given too but I want to know that can it be done using combinations and the method I have explained in part 2 above. What is wrong in arranging the way it is done. That way ans would be wrong but what i am trying to understand is error in fundamental in part 2 stated by me.

3c1 x 2c1 X 1c1 X 2c1 X 1c1 = 12

ans then arranging it is 5! WAYS i.e. 120 X 12

[sumit82]

Let me put another question that might explain what I am trying to say..

Q1. How many 4 digit numbers can be formed using 1,2,3,4,5,6,7,8,9.

Ans - 9c1 X 8c1 X 7c1 X 6c1

Q2. How many 4 digit numbers can be formed using 1,1,2,2,5,6,7,7,9.

Ans - Plz solve


Cheers

[sunil_dr]

Originally Posted by newkid

answer can be obtained using permutations

it is 5!
3! 2!

total no. of steps is 5
3 r of one kind and 2 are of other kind

hence the result.

Hey Thanks! This really helps. I was using the OG method of listing all combinations, which is pretty time consuming and error prone.

[newkid]

Q 1. given nos. 1,2,3,4,5,6,7,8,9


now, let me make 2 things clear to u.
first, every problem has to be taken with the precise information. now, the problem raised by u does not state anything about repitions. i will clarify the position with examples as shown below.

second, u have to understand the basic difference between permutation and combination. symbolically permuation is nPr = n! divided by (n-r)! . next combination is nCr which is given by n! divided by the product of r! and (n-r)! .

so, if u carefully analyse, the differnce is the extra division by r! in case of combinations. that is so because, in permuations ( same as arrangement) the interse arrangement of the selected class of items ( i.e., represented by r here) is important. However, in combination , the interse arrangement is not taken into account and it is simply the group of r items without any weight assigned to the interse arrangement.

Before answering Q1, i once again state that the city A - B problem is a permutation problem and not a combination problem. three steps up and 2 step right have to be looked vis a vis the order of steps ( i.e. urruu or say uuurr or even say rruuu) . Had it been a combination problem , with the restrictions of minima, the anwer in an infinte grid would also be one (1) only. But, since the order of the steps is vital , it becomes a permutation problem.

coming to Q 1,
A) if repitions are allowed, the answer would be 9 raised to the power of 4 (i.e., 9 x 9 x 9 x 9)

B) if repitions are not allowed, that is , one no. can be used only once, it is 9 P 4

i) 9 p 4 = 9! divided by (9-4)!
ii) alternatively, using simple logic, the answer can be derived as :- for the Ist place , any of the 9 nos. can be taken. Next, for the 2nd place, any of the 8 remaining nos. can be taken. Similarly, for the 3rd and 4th place any of the remaining 7 and 6 nos. can be taken.
Result 9x8x7x6

Both the approaches (i) and (ii) leads to the same answer. (i) is based on direct formula and (ii) is based on logical solution

coming to Q 2 ............

[newkid]

coming to question 2. the nos are say 1,1, 2,2, 3, 4, 7,7,9

we have 9 nos.

the total no of different groups is 6
three groups have two elements each which are same.

now. suppose one of the nos. is say 1123. ( assuminng no repitions)
now interchanging digits at hundredth and thousandth position we still get the no. as 1123. this is so because of both the nos. being same. Hence, in such cases we have to take into account such repitions of the elements in the problem as given to arrive at the correct solutions.

had all nos been different, answer would have been 9x8x7x6 = 3024. but in these case we have 3 groups with 2 elements each the answer would be as under

( 9x8x7x6 ) / (2! 2! 2!) = 3024 / 8 = 378.

if it is still not clear, i give u a more simple example

say we have nos. 1 2 3 and to make say, 2 digit nos.

answer would be 3x2 = 6 -------> ( 12 , 13, 21, 23, 31,32)

Now, assume the given nos are say 1, 1 and 2

the answer will be modified as under

( 3x2 ) / 2! = 6/2 = 3 ------> ( 11, 12, 21)


RIGHT.

i think , now u can follow the logic.
:whatthat: :grab: