Going To New Year Parties

There are five parties, which I will label: - A (Andrew) - B (Betty) - C (Chandra) - D (Dela) - E (Eugene)

We can represent the time slots of the five parties as follows:

X, X, X, X, X

Since Eugene's party will always be last, we can immediately fill in the last slot:

X, X, X, X, E

Before we apply the other condition, in which Andrew's party is attended before Betty's party, let's determine how many ways we can place the four parties in the remaining 4 slots. This would be $4 \times 3 \times 2 \times 1= 4! = 24$ . Note that if we apply the last condition in which A comes before B, we need only to divide by 2 because we can interchange A and B's positions to get a new instance, so if we divide by 2 we fix the instance. For example:

C, A, D, B, E and C, B, D, A, E are possibilities in which A and B are interchanged. By dividing by two we force only one of these interchanges to be counted. Thus, $24 / 2 = \boxed{12}$

Here are all the possibilities: ABCD, ABDC, ACBD, ADBC, ACDB, ADCB, CABD, DABD, CADB, DACB, CDAB and DCAB. So in total there are 12 possibilities!

We can represent it like this _ _ _ _ E

Total ways 4!

In half of them

Andrew's party is before Betty's party

And in other half Andrew's party is after Betty's party

required answer 4!/2=12

Keeping A first possible arrangments are 6 then keeping C and D first total 6 arrangments are possible

3P3 = For the three "favored" and "scheduled" parties 2P2 = For the remaining parties

3P3*2P2 = 12

Just put Eugene last, so you have got 4 places and so 24 choices but in half of these choices Andrew's party comes before Betty's and in the other half Betty's party comes before Andrew's. So the answer is 24/2=12

This problem is analogous to arranging five letters (say, A,B,C,D,E) to form five-letter words. By comparison, the last letter E is fixed, since the guy wants to attend Eugene's party last in all cases. There are remaining 4 letters and 4 slots. There are 4! possible arrangements. But, it is given that the guy wants to go to Andrew's before Bella's. That is, the order of the letters A and B is invariant. So we replace the letters A and B by a common letter X. Hence, the number of arrangements = 4!/2! = 12.

Let the Places be A,B,C,D and E.

Given E must be at Last.

So we need to arrange A,B,C and D such that B comes only after A.

The Probability that B comes after A is 1/2.

The Total no: of ways we can arrange A,B,C and D are 4!

Of those we need B appearing after A.So only half of it would be the the no: of chances.

Hence 12.

3!+2.2!+2!=12

For the sake of convenience, I let A represent Andrew, B represent Betty, C represent Chandra, D represent Dela and E represent Eugene. The key to solving this problem is that E must always be at the back of the order. Also, the person must go to A before B.

Also, for any of the 6 variations, simply interchange C and D to obtain 12 ways.

The 12 possible ways are:

1) ABCDE 2) ABDCE 3) ACBDE 4) ADBCE 5) CABDE 6) DABCE 7) ACDBE 8) ADCBE 9) CADBE 10) DACBE 11) CDABE 12) DCABE

Since there are only four people in the picture thus there are 12 possible ways |: - )>

ABCDE - ACDBE - ACBDE

ABDCE CDABE DCABE For each case there are 3 different ways. So 3X4 = 12

toal 24/2=12