P n C

2) There are 5A's , 5B's , and 5C's. In how many ways they can be arranged in 15 boxes such that first five boxes do not contain any A, middle five do not contain any B and the last five do not contain any C ??

1) Straight lines are drawn by jining 'm' points on a straght lne to 'n' points on another line. Excluding the given points(m+n), find the number of intersection points if no two drawn are parallel and no three are concurrent.

Well to get one point of intersection we need 2 points on the first line and 2 points on the second line.

Now 2 points on the first line can be selected out of m points in ^mC₂ ways.

Similarly we can select 2 points on the second line out of n points in ⁿC₂.

Therefore the required n. of lines = ^mC₂ x ⁿC₂
= (1/4) mn (m-1)(n - 1)

2) There are 5A's , 5B's , and 5C's. In how many ways they can be arranged in 15 boxes such that first five boxes do not contain any A, middle five do not contain any B and the last five do not contain any C ??

This can be done, as I think,

Since first 5 boxes should not contain any A's, we have to fill them using only 5 B's and 5 C's.
They may contain all the C's only or B's, they may contain 2 C's and 3 B's or vice versa, they may contain 1 C and 4 B's or other way round ,i.e, 1 B and 4 C's. So the letters can be arranged in them in following ways,

5P0 x 5P5/5! + 5P1 x 5P4/4! + 5P2/2! x 5P3/3! + 5P3/3! x 5P2/2! + 5P4/4! x 5P1 + 5P5/5! x 5P0

= 252.

I divided above terms because am assuming all the A's are identical, similar assumptions for B and C also.

We can do the same kind of arrangement for next two set of 5 boxes excluding all the B's for second 5 boxes and all the C's for the last set of 5 boxes.

So the total no. of arrangements can be = 252 + 252 + 252 = 756.

2nd one i think u have included the cases many times...

for example if first five have 2B and 3C, then how can next five have 3C or 4C or 5C????

think.....