A group of X families, each with Y members are to be lined up in a sin

CEdward
Since each family has Y! members, whatever ways the members are arranged they have to be arranged for each of the family.
Hence Y! for 1st family then Y! for the 2nd family and Y! for the 3rd family and so on...

Try plugging in integers for X and Y.
Example 1: X = 2, Y = 3
Let's say A,B and C are members of family 1 and D, E and F are of family 2, and each of these family members can be arranged within themselves in 3! ways + each of the families can be arranged in 2! ways.
Thus, Number of ways 6 persons can be arranged are
ABC*DEF
\(3!*3! = 3!^2\)

But there's another possibility which differs from above arrangement i.e.
DEF*ABC
\(3!*3! = 3!^2\)

Hence, Total ways = \(2*3!^2= 2!*3!^2 = X!*Y!^X\)

However, you need to check for other integers as one set of integers can satisfy 2 or 3 answer choices.

Example 2: X = 3, Y = 4
Let's say A,B, C and D are members of family 1; E, F, G and H are of family 2 and I,J,K and L are of family 3, and each of these family members can be arranged within themselves in 4! ways + each family can be arranged in 3! ways.
Thus, Number of ways 12 persons can be arranged are
ABCD*EFGH*IJKL
\(4!*4!*4! = 4!^3\)

But there's a possibility which differs from above arrangement i.e.
EFGH*ABCD*IJKL
\(4!*4!*4! = 4!^3\)

AND there's another possibility which differs from above arrangements i.e.
IJKL*ABCD*EFGH
\(4!*4!*4! = 4!^3\)

and so on.... i.e. total 6 ways.

Hence, Total ways = \(6*4!^3 = 3!*4!^3 = X!*Y!^X\)

Therefore answer is D.

A group of X families, each with Y members are to be lined up in a single row for a photo.
How many ways can the people be arranged if each family member must be kept together?

The number of ways to arrange X families = X!
The number of ways to arrange Y members of a family = Y!
The number of ways to arrange Y members for X families = \(Y!ˆX\)

The Number of ways the people be arranged if each family member must be kept together =\( X!* Y!ˆX\)

IMO D
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