1.9c3*3!=504

2.8c4*4c2=402

If these are correct, i will explain my reasoningn later. I'm in a hurry at work.

(9C3 x 6C3 x 3C3)/3! is the number of ways of splitting 3 students into three different groups
Since they are going to tackle 3 different problems
((9C3 x 6C3 x 3C3)/3! ) x3! = 1680

Do we have an OA on these?

1. 9!/3!3!3! = 1,680
2. 8!/2!2!2!2! = 2,520

May be I'm wrong?!

erdnah wrote:1. 9!/3!3!3! = 1,680
2. 8!/2!2!2!2! = 2,520

May be I'm wrong?!

nope .. ur right ..

erdnah or gabriel, can you explain?

Ok .. this is the question ..

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

we have to divide the 8 alaskan huskies into 4 groups of 2 each ..

the first 2 can be selected in 8c2 ways ..

the second group can be selected in 6c2 ways ( 6 bcoz that is the number of huskies left after the first group is chosen) ..

similarly the 3rd group can be chosen in 4c2 ways..

the last group can be chosen in 2c2 ways ..

so the total no. of ways in which the 8 alaskan huskies can be divided into 4 groups of 2 each is = 8c2*6c2*4c2*2c2 .. which is basically equal to 8!/( 2!*2!*2!*2!) ..

PS : if the problem asked us to only divide the 8 huskies into 4 groups then we would have to divide the answer by 4! .. so the evntual answer wuld be 8!/( 2!*2!*2!*2!*4!) ..

but over here the question asks us to divide the 8 huskies into 4 groups each of which will be assigned to a different sled .. so we wont divide the answer by 4! ..