Chairs

[ricky]

Can you help me on this?

[2008]

maybe is wrong, but as with 2 chairs out of 5 you get 10 possible combinations and the total combinations are 150, then the combinations of tables must be 15... only 6 gives this possibility.

it took me a while to get to this, any fast solution?

[ricky]

Why cant the number of chairs selection be 20? P(5,2).
First chair can be selected in 5 ways and second can be selected in 4 ways.Therefore Total = 5*4=20!...

[2008]

mmm i m a new, so i m not that confident, but i though:
it says there are 5 chairs, example: A B C D E.
possible selections
AB AC AD AE
BC BD BE
CD CE
DE

i think...

ps. by the way, ricky stands for riccardo? i m looking for someone in milan to study with...

[aspire750]

Since a total of 150 combinations are there. It has to be
150=No. of combinations of chairs* no. of combinations of tables
No. of ways chairs can be arranged=5C2=10
Therefore, no. of ways tables can be arranged would be 15.
6C2=15. Hence, there has to be 6 tables.

[ricky]

I understand everything but why cant be there 20 possible selections of chairs.
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?

[2008]

I understand everything but why cant be there 20 possible selections of chairs.
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?

first chair (A) can be selected with 4 other chairs: AB AC AD AE
second chair (B) can then be selected in 3 other chairs BC BD BE
third chair (c) can be selected in 2 more ways: CD CE
last two chairs can be selected togher: DE

4+3+2+1=10

[parallel_chase]

I think the answer is 6 it can never be 30

no. of chairs = 5
no. of tables = x

5C2 * xC2 = 150

10 * xC2 = 150

xC2 = 15

6C2 = 15

Hence the answer is 6.

GMAT prep software is known for having bugs such as these.

[lalitgmat]

Please See the question does NOT mention if their chairs and Tables are unique to each other. Hence, we can NOT name chairs as A,B,C,D..
& 5C2 * NC2 = 150.
=> N =6

[Stuart@KaplanGMAT]

I understand everything but why cant be there 20 possible selections of chairs.
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?

Because you have duplicates - the way you counted, chair A then chair B is a different selection than chair B then chair A - when in fact those selections are identical.

So, you could do the calculation (5*4)/2 = 20/2 = 10

[arorag]

HI Stuart,

Please can you explain the concept of deplicates in per./com./prob. with some example.
Thanks in advance

[Stuart@KaplanGMAT]

HI Stuart,

Please can you explain the concept of deplicates in per./com./prob. with some example.
Thanks in advance

Duplicates occur when you count the same arrangement more than once.

In combinations, this happens when we forget that order doesn't matter.

For example, let's say I want to choose a pair of people out of 4 possibilities, A B C and D.

If I were to say "well, there's 4 possibilities for the first person and 3 possibilities for the second person, so there's 4*3 = 12 possible pairs", I'd be double counting because I'm forgetting that the pair "AB" is the exact same as the pair "BA".

In permutations, duplicates happen when we have identical entities, i.e. not all the entities are unique.

For example, if I want to know all the possible ways to arrange the letters in the word "GMAT", the answer would simply be 4!.

However, if I want to know all the possible ways to arrange the letters in the word "KAPLAN" and use the same approach, I'd get an answer of 6!, which would be wrong, since "KAPLAN" has two As.

Let's call the As in Kaplan A(1) and A(2) and see what I mean.

KA(1)PLA(2)N and KA(2)PLA(1)N look exactly the same, but we've counted it as two separate arrangements.

In order to avoid duplications in permutation questions, you need to factor out the number of duplicates.

So, the true number of arrangements of "KAPLAN" is 6!/2!.

Let's look at some other examples:

DESSERT... 7!/2!2!, since we need to factor out 2 Es and 2 Ss.

DESSERTS... 8!/2!3!, since we need to factor out 2 Es and 3 Ss.

DESERTERS... 9!/3!2!2!, since we need to factor out 3 Es, 2 Ss and 2 Rs.

The basic formula is n!/r!s!t!... in which n is the total number of items and r, s, t, ... are the number of duplications.