We need to find the number of arrangements of Father, Mother, Child-1, Child-2, and a Dog, so that F and M are always together.

Consider F and M to be one unit, then we would have total of four units: {FM}, {C1}, {C2}, {D}. The number of arrangements of these four units is 4!, but F and M within their unit can be arranged in 2 ways: {FM} or {MF}, so total number of arrangements is 4!*2=48.

Answer: C.
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F, M, C1, C2, D (let those be Father, Mother, Children 1, Children 2 and Dog respectively)

We need to find all the variations for:

F, M, X, X, X --> the Arrangement for the three spots is 3!, so 6 options. Also need to consider the other 6 options when mother is at position one and father at 2. So in total 12.
X, F, M, X, X --> the same
X, X, F, M, X --> the same
X, X, X, F, M --> the same

The total number of variatons would be 12*4, which is 48.

Total people: \(F, M, C_1, C_2, D\) [Row arrangement]

Condition: Father insists on standing by his woman (F, M) forms 1 group

=> C_1, C_2, D, (F, M) = arranging 4 people will take 4! ways and F, M can be arranged in 2! ways.

Total ways: 4! * 2! = 24 * 2 = 48

Answer C
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Family has five members. Two members (mother&father) must be next to each other. So, we should consider them as one member, who can be arranged in two variations (mother&father; father&mother). Now we have four members and four spots for them __ __ __ __. The first spot can take 4 members, the second - 3, the third - 2, and the last - 1. It will be 4*3*2*1=4!=24 arrangments, but we should remember that our "unique" member (which includes mother&father) can be in two variations, so 24*2=48.

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