How to pool variables that rarely occur, particularly with respect to survey data From the text : Multiple Correspondents Analysis by Brigette LeRoux

very infrequent categories of active variables need to be pooled with
  others when feasible

The text doesn't explain how this should be done, and I'm finding it hard to find literature / examples of this being done.
For example, if I have variables x1,x2 as follows
x1 = [1,1,1,1,0,0,1,0,1,1,1,1,0,0,1,0,0,1,1,0]
x2 = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
x3 = [0,1,0,1,0,0,1,0,1,0,0,1,0,0,1,1,0,0,0,1]

Then I might want to pool x2, but how would I go about doing that?
Edit
"other". So, if you surveyed people about their favorite colors and 45 % said "pink" and 45% said "blue" and you had many other answers that were infrequent, you might pool the rest into "other". – Sal Mangiafico 41 mins ago 
Following an example from @SalMangiafico in the comments:
Suppose that I have two variables pink (p) and fuchsia (f) as follows
p = [1, 1, 1, 0]
f = [1, 0, 0, 1]

How would they be combined? Would be be using an or operation such as
p or f = [1, 1, 1, 1]

or are there other ways to approach this?
 A: Let's start with a concrete example.
Let's say I survey people on a happiness index.  And in the survey I also ask for their favorite color.  I want to see if their favorite color can predict happiness.  I use dummy coding for the favorite color variables.
Here, most people answer "Pink" or "Blue", but there are also rare answers of "Fuchsia" and "Black"
(The following code can be run in R or at https://rdrr.io/snippets/ ).
Happiness = c(10, 11, 12, 13, 15, 14, 20, 25, 30, 32, 100, 20)
Pink     = c( 0,  0,  0,  0,  0,   1,  1,  1,  1,  1,   0,  0)
Blue     = c( 1,  1,  1,  1,  1,   0,  0,  0,  0,  0,   0,  0)
Fuchsia  = c( 0,  0,  0,  0,  0,   0,  0,  0,  0,  0,   0,  1)
Black    = c( 0,  0,  0,  0,  0,   0,  0,  0,  0,  0,   1,  0)

Data = data.frame(Happiness, Pink, Blue, Fuchsia, Black)

Option 1:  Ignore rare colors.  
Here, I'll ignore both Fuchsia and Black.  I will make a new data set with these observations totally excluded so that they are not combined by the software with either Pink or Blue.
Data1 = Data[Data$Pink==1 | Data$Blue ==1,]

model1 = lm(Happiness ~ Pink, data=Data1)

summary(model1)

   ### beta = 12.0, p = 0.008, r-squared = 0.609

Option 2: Combine rare variables.  
Another option is to combine rare values either with other rare values or with a common value that is relevantly similar.  For example, we could combine Fuchsia and Black into a variable "other".  
Instead, in this case, let's say Fuchsia is close enough to Pink to combine them into one variable.
Note here that the data are dummy coded as 0 (false) and 1 (true), so that it makes sense to use addition to combine them (1 + 0 = 1; 0 + 1 = 1; 0 + 0 = 0).  If the values were TRUE and FALSE, we could use the Boolean operator OR to combine them (TRUE | FALSE = TRUE, FALSE | TRUE = TRUE, FALSE | FALSE =FALSE).
Also note that it's important to be sure that the values are coded in the way you expect.  Here, I'm treating 1 as TRUE, but if the data were coded where 1 were FALSE, I would use a different operation to combine the variables.
Data$PinkFuchsia = Data$Pink + Data$Fuchsia

Data$PinkFuchsia

   ### [1] 0 0 0 0 0 1 1 1 1 1 0 1

Because the Happiness value for the Fuchsia observation isn't terribly different than those for Pink, the model doesn't change much, and we can feel relatively justified in combining these variables.
Data2 = Data[Data$PinkFuchsia==1 | Data$Blue ==1,]

model2 = lm(Happiness ~ PinkFuchsia, data=Data2)

summary(model2)

   ### beta = 11.3, p = 0.006, r-squared = 0.586

Option 3:  Include the rare Black variable as it is.
Data3 = Data[Data$Pink==1 | Data$Blue==1 | Data$Black==1 ,]

But note here that including the one Black observation changes the model greatly. We would want to be cautious here.  Are we allowing a single (or rare) observation to matter too much in our results? Because the values are rare, it's difficult to know how to answer this question, and we might want to avoid this approach.
model3 = lm(Happiness ~ Pink + Black, data=Data3)

summary(model3)

   ### For Black, beta = 87.8, p = < 0.0001. r-squared for model = 0.965

