Choose 2 juniors and 2 seniors.

14C2*23C2 = 23023

The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.

13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second Or

Case II: 4 students in the first car and 4 in the second

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is

8C3+8C4 = 56 + 70 = 126.

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

There are 5 slots.

                   __ __ __ __ __

The first slot must be a four. There are 4 ways to put a four in the first slot.

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

(4)(4)(4)(4)(4) = 1024

Therefore there are 1024 different ways to produce the desired hand of cards.