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# Four families of three are lining up for a photo. How many ways can th

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Math Expert
Joined: 02 Sep 2009
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Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 00:26
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Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?

A. $$12$$

B. $$4!3!$$

C. $$7!$$

D. $$4!6^4$$

E. $$12!$$

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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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Updated on: 05 Dec 2019, 04:53
1
no.of ways of arranging 4 families is in 4! ways
no.of ways of arranging 3 members in 4 families is in 3!^4= 6^4 ways
Total ways= 4!(6)^4

OA:D

Originally posted by madgmat2019 on 05 Dec 2019, 01:30.
Last edited by madgmat2019 on 05 Dec 2019, 04:53, edited 1 time in total.
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Four families of three are lining up for a photo. How many ways can th  [#permalink]

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Updated on: 14 Dec 2019, 10:52
1
No. of families, n=4
size of a family=3.
Since each family are to be together and not separated, we can represent the families with a,b,c, and d.
so we have the following possible aaa, bbb, ccc, ddd. Basically, we cannot distinguish between the members of the families. So we have abcd. We can arrange four (abcd) families in 4! ways
Now each of the three families can have their three members arranged in 3! ways

Hence total number of ways = 3!*3!*3!*3!*4! = 4!*3!^4

The right answer is option D.

Originally posted by eakabuah on 05 Dec 2019, 02:16.
Last edited by eakabuah on 14 Dec 2019, 10:52, edited 1 time in total.
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Posts: 32
Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 02:25
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Since there are 4 families and each family needs to stand together, let's assume 1 family as one block. Thus, there are total 4 blocks and each block has 3 people.

These families (4 blocks), can arrange themselves in 4P4 = 4! ways.

Each family(block) has 3 members. Now, wherever a family stands, they can arrange themselves in 3! ways.

For example, if a family has X, Y, and Z as three members...these three can arrange themselves in 3P3 = 3! ways.

Total arrangement = $$4! *3!*3!*3!*3! = 4! *6^4$$

The correct answer is Option D.
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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 03:27
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Quote:
Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?

A. 12

B. 4!3!

C. 7!

D. 4!6^4

E. 12!

Families = 4: [ABCD] = 4! arrangements
Each Family has 3 members: 3! arrangements
Total arrangements: $$4!*3!^4=4!6^4$$

Ans (D)
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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 03:30
1
Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?

A. 12

B. 4!3!

C. 7!

D. $$4!6^4$$

E. 12!

Each family makes a row so there are four rows. In each row the members can be arranged in 3! ways
So possibilities for various rows are as follows:
1st - 4 ways(families)
2nd - 3 ways
3rd - 2 ways
4th - 1 way

Hence total possibilities = $$4! * 3! * 3! * 3! * 3! = 4! * (3!)^4$$

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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 03:39
I think it's (B).
Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?
Treat each family as one --> we have 4! ways to arrange the four families within the line
Within each family --> members can be arranged in 3! ways
For a total: 4! x 3!
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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 03:40
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4 families can be arranged in 4! ways and within each family total ways ; 3! ;6 for each of 4 ways the families can be arranged in 6^4 ways
IMO D; 4!*6^4

Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?

A. 1212

B. 4!3!

C. 7!

D. 4!6^4

E. 12!
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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 11:06
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Q - Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?

Lets consider each family of three as single unit, so 4 families = 4 items, so they can be arranged in 4! ways.
Now, within each families, 3 members can further be arranged in 3! ways.
Thus, number of they can be lined up in 4! * 3!*3!*3!*3! ways(3! for each 4 families)
After simplifying 4! * 6^4.
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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 13:27
4 families can line up in 4! ways. Each family has 3 members. So each family can line up in 3! ways. So the numer of ways are 4!*3!.

I'm wondering if one family has 2 members, another has 4 members, and the remaining has 3, what will be the answer?
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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 19:06
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4 families, each of which has 3 members. Members of the family must stay together.

Ways that 4 families line up with exactly same arrangement of their members= 4!

Within each family, its 3 members can swop their place
= 4!*(3!*3!*3!*3!)
= 4!*(6^4)

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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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05 Dec 2019, 22:29
The four families can arrange themselves in 4! Ways ABCD, BCDA.

The families comprise of 3 members they can arrange themselves in 3! Ways.

Total ways 4!*3!
IMO B

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Re: Four families of three are lining up for a photo. How many ways can th  [#permalink]

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06 Dec 2019, 01:03
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The # of possibilities of Four families —> 4!

—> each family has 3 members —> 3!*3!*3!*3!

The total possibilities —>

$$4!*3!*3!*3!*3!= 4!*6*6*6*6= 4!*6^{4}$$

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Re: Four families of three are lining up for a photo. How many ways can th   [#permalink] 06 Dec 2019, 01:03
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