old_dream_1976 wrote:

sharadGmat wrote:

I am posting the good ones for your practice..

4. Eight Alaskan Huskies are split into pairs to pull one of four sleds in a race. How many different assignments of Huskies to sleds are possible?

(A) 32

(B) 64

(C) 420

(D) 1680

(E) 2520

good question but is it not necessary to specify that only 2 dogs pull 1 sled?

if all 8 pull 1 sledge and if pairs are of importance then there would be

7 x 5 x 3 x 4 = 420 possibilities.

Can the math Gods confirm this ?

I think problem is more complicated.. The different teams possible =

(8C2*6C2*4C2*2c2)/4! .. Here we need to do 4! for the following reasons..

When we divide a,b,c,d into 2 teams there could be 4c2*2c2*=6 ways..

ie.

1. ab,cd

2. ac,bd

3 ad,bc

4. bc,ad same as 3

5. bd,ac same as 2

6. cd,bc same as 1

So we need to divide 6/2! to get the actual number of ways.. ie. 3.

Similarly for 4 teams , we need to divide b y 4!.

So in our case, there are total = (8C2*6C2*4C2*2c2)/4! =105 ways..

So these teams-ways or say team-sets be named as 1, 2, 3, 5 through 105 where each one has 4 teams..

Each team-set has 4 teams.. And these need to be assigned to 4 sleds..

So total ways = 105* 4(number of teams) P4(number of sledges) = 105*4!= 2520..

So in your case, if all the 8 dogs pull one sledge, then

105*4P1 = 105 * 4 = 420..

Making sense ?