Order Santa's Reindeer

Use combinatorial rule to calculate the way for ordering the reindeer.

The first two reindeer can be ordered by using $C(8,2)= 28$

The second two reindeer can be ordered by using $C(6,2)= 15$

The third two reindeer can be ordered by using $C(4,2)= 6$

The last two reindeer can be ordered by using $C(2,2)= 1$

The total ordering of reindeer is: $28 * 15 * 6 * 1= 2520$

So, the last three digits is 520

For the first row, there are $8$ reindeers and we must choose $2$ of them. The number of possible ways of choosing $2$ reindeers from $8$ is given by the binomial coefficient, ${8 \choose 2} = 28$ We continue to use the same process for the next rows. For the second row we have to choose $2$ reindeers from $6$ , for the third row we have to choose $2$ reindeers from $4$ and for the final row we have to choose $2$ reindeers from $2$ Hence, by the product rule the number of combinations is given by, ${8 \choose 2}{6 \choose 2}{4 \choose 2}{2 \choose 2} = 2520$ Therefore, the final answer is $\fbox{520}$ .

since the ordering in each row is inconsequential, we must use simple combination. so ,Santa first chooses 2 from 8 reindeer and then another 2 from 6 reindeer ,2 from the remaining 4 and finally the last 2 ... So, the answer is 8C2 x6C2x 4C2x2C2=2520........ Thus the last 3 digits are 520 MERRY CHRISTMAS TO ALL!!!!

According to the question, they form two lines behind Rudolph of 4 in each. Now keeping it in mind that left-right ordering within a row does not matter, we get the answer simply by ${8 \choose 2 }$ $\times$ ${ 6 \choose 2 }$ $\times$ ${ 4 \choose 2 }$ $\times$ ${ 2 \choose 2 }$ that is 2520. so our answer is 520.

$8\choose2$ $6\choose2$ $4\choose2$ $2\choose2$ = $2520$

So the last three digits are $\boxed{520}$

8!/4.(8/2) = 8!/4.4 = 16520 ---> Answer : 520