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by gmatusa2010 » Tue Dec 28, 2010 6:40 pm
Out of seven models, all of different heights, five models will be chosen to pose
for a photograph. If the five models are to stand in a line from shortest to tallest
and the fourth-tallest and sixth-tallest cannot be adjacent, how many different
arrangements of five models are possible?
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by anshumishra » Tue Dec 28, 2010 7:01 pm
gmatusa2010 wrote:Out of seven models, all of different heights, five models will be chosen to pose
for a photograph. If the five models are to stand in a line from shortest to tallest
and the fourth-tallest and sixth-tallest cannot be adjacent, how many different
arrangements of five models are possible?
Choose 4 and 6, we must also choose 5 who will stand between them =3C3*4C2=4C2=6
Choose either 4 or 6 = 2*1*5C4=10
Choose neither 4 nor 6 = 5C5=1

So, total no. of arrangements = 17.
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by gmatusa2010 » Tue Dec 28, 2010 7:12 pm
wow. can you expand a little bit here?
anshumishra wrote:
gmatusa2010 wrote:Out of seven models, all of different heights, five models will be chosen to pose
for a photograph. If the five models are to stand in a line from shortest to tallest
and the fourth-tallest and sixth-tallest cannot be adjacent, how many different
arrangements of five models are possible?
Choose 4 and 6, we must also choose 5 who will stand between them =3C3*4C2=4C2=6
Choose either 4 or 6 = 2*1*5C4=10
Choose neither 4 nor 6 = 5C5=1

So, total no. of arrangements = 17.
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by anshumishra » Tue Dec 28, 2010 7:21 pm
gmatusa2010 wrote:wow. can you expand a little bit here?
anshumishra wrote:
gmatusa2010 wrote:Out of seven models, all of different heights, five models will be chosen to pose
for a photograph. If the five models are to stand in a line from shortest to tallest
and the fourth-tallest and sixth-tallest cannot be adjacent, how many different
arrangements of five models are possible?
Choose 4 and 6, we must also choose 5 who will stand between them =3C3*4C2=4C2=6
Choose either 4 or 6 = 2*1*5C4=10
Choose neither 4 nor 6 = 5C5=1

So, total no. of arrangements = 17.
Sure.
There are only 3 possible cases. We choose both 4 and 6, Either 4 or 6 , Or Neither 4 nor 6

Choose 4 and 6, we must also choose 5 who will stand between them =3C3*4C2=4C2=6 {3C3 -> No. of ways of choosing 4,5 and 6 out of 4,5 and 6. 4C2-> No. of ways of selecting the remaining 2 out of 4 left}

Choose either 4 or 6 = 2*1*5C4=10 {No. of ways to select either 4 or 6 is 2*1 or just say 2, 5C4 -> choose the remaining 4 from the other 5 left}

Choose neither 4 nor 6 = 5C5=1 {If you choose neither 4 nor 6, you are left with remaining 5 models. So number of ways of selecting 5 models out of 5 = 5C5 or just say 1}

So, total no. of arrangements = 17.

Let me know if you have any doubts.[/i]
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Anshu

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by gmatusa2010 » Tue Dec 28, 2010 7:33 pm
I still dont understand how 1) You are restricting the two (4 and 6) from being next to each other 2) You are restricting the fact that the people have to be from shortest to tallest.
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by anshumishra » Tue Dec 28, 2010 7:47 pm
gmatusa2010 wrote:I still dont understand how 1) You are restricting the two (4 and 6) from being next to each other 2) You are restricting the fact that the people have to be from shortest to tallest.
Here we are arranging people (Think about permutation, right ?), BUT there is ONLY ONE WAY to arrange the selected models , in order of height, so it reduces to just selection (Combination) problem. This answers your 2nd question.

I am restricting 4 and 6 from being next to each other just by selecting 5, whenever I select 4 and 6. Just selecting them together ensures that they don't stand side by side, because there is only one way of arranging them, in order of height.
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