Question

Ten members of a wedding party are lining up in a row for a photograph. (a) How many ways are there to line up the ten people if the bride must be next to the maid of honor and the groom must be next to the best man? (b) The photographer wants pictures of all possible groups of 4. How many pictures must the photographer take? (c) The photographer now wants to select 3 of the men (of which there are 6) and 3 of the women (of which there are 4) to be in a picture. How many possible ways could this be accomplished? (d) Now 22 people are present for the pictures. 5 are family members for the groom (FG), 2 are family members for the bride (FB), 4 are groomsmen (GM), 4 are bridesmaids (BM), and 5 are friends (F). Finally, there are the (B)ride and (G)room. The photographer is trying to determine how many ways to arrange them based only on their relationship to the wedding couple and not based on their individual identities. Help the photographer determine how many arrangements exist.

Answer 1

(a) We want to line up the members such that bride is always next to maid of honour and groom next to best man. So we consider the bride and maid of honour to be one unit and groom and best man to be one unit and rest of the individuals to be one unit each. We have 8 units which can be permuted in 8! Ways and 2 units of two people each can be permuted in 2! Ways within themselves. So total number of ways = 8!2!2! =161280

(b) Any group of 4 from 10 members can be selected in ways so the number of pics that the photographer has to take is also the same that is :

(c) Number of ways of selecting 3 men out of 6 =

   Number of ways of selecting 3 women out of 4 =
  

So total number of ways of selecting this combination=
     

(d) Now we have 22 people and we want them to be arranged according to relationship to the wedding couple so we make one unit of the groom and one unit of the bride so that the people related to groom always stay together in the picture and people related to the bride always stay together. We let the 5 friends to be neutral individual units (as no relationship is given in question)

So unit of groom = (FG + GM + G) = 5 + 4 + 1 = 10

So unit of bride = (FB+ BM + B) = 2 + 4 + 1 = 7

And five individual units

So first we permute the units in 7! Ways as there are 7 units and then we permute within the B unit in 7! Ways and within the G unit in 10! Ways

So total arrangements possible = 7!7!10! = 9,21,77,32,60,80,000

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