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# Out of seven models, all of different heights, five models

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Director
Joined: 11 Jun 2007
Posts: 595
Out of seven models, all of different heights, five models [#permalink]

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Updated on: 17 Feb 2012, 07:33
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Question Stats:

56% (02:09) correct 44% (02:09) wrong based on 475 sessions

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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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Originally posted by eyunni on 27 Oct 2007, 12:34.
Last edited by Bunuel on 17 Feb 2012, 07:33, edited 1 time in total.
Edited the question and added the OA
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Posts: 47015
Re: Out of seven models, all of different heights, five models [#permalink]

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17 Feb 2012, 07:42
10
5
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Notice that since the fourth-tallest and sixth-tallest models cannot be adjacent and models are to stand in a line from shortest to tallest, then if we choose 4th and 6th we MUST also choose 5th to stand between them.

Also notice that since the models are to stand in a line from shortest to tallest then for any group of 5 models we choose, there will be only one arrangement possible: from shortest to tallest.

Hence we can have 3 cases:
Choosing 4th, 5th, and 6th tallest models and any 2 from 4 models left: $$C^3_3*C^2_4=6$$;
Choosing either 4th or 6th tallest models and any 4 from 5 models left: $$C^1_2*C^4_5=10$$;
Choosing neither 4th nor 6th tallest models: $$C^5_5=1$$;

Total: 6+10+1=17.

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Re: Out of seven models, all of different heights, five models [#permalink]

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20 May 2012, 04:19
9
1
Total no. of ways "5 models can be selected from 7" = 7C5 = 21.

Fourth tallest and Sixth tallest can never stand adjacent to each other. Also, all selected models will be lined up from shortest to tallest.
Hence, we have to pull out the no. of ways 4 & 6 stand adjacent to each other from the above total, i.e = 21.

The following will be the way in which 4 & 6 stand close to each other.

1 2 3 4 6
1 2 4 6 7
1 3 4 6 7
2 3 4 6 7

So, the answer will be 21-4 = 17 ways.
##### General Discussion
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Re: Out of seven models, all of different heights, five models [#permalink]

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18 Feb 2012, 14:09
1
Bunuel wrote:
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Notice that since the fourth-tallest and sixth-tallest models cannot be adjacent and models are to stand in a line from shortest to tallest, then if we choose 4th and 6th we MUST also choose 5th to stand between them.

Also notice that since the models are to stand in a line from shortest to tallest then for any group of 5 models we choose, there will be only one arrangement possible: from shortest to tallest.

Hence we can have 3 cases:
Choosing 4th, 5th, and 6th tallest models and any 2 from 4 models left: $$C^3_3*C^2_4=6$$;
Choosing either 4th or 6th tallest models and any 4 from 5 models left: $$C^1_2*C^4_5=10$$;
Choosing neither 4th nor 6th tallest models: $$C^5_5=1$$;

Total: 6+10+1=17.

Brilliant, had the same idea, but didn't know how to model it with combinations.
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Re: Out of seven models, all of different heights, five models [#permalink]

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24 May 2012, 11:12
1
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

You could also solve it in this way -

Since fourth and sixth tallest cannot be adjacent, the only way for both of them to be in the chosen group and not stand adjacent is to have the fifth tallest model stand between them , as the models have to stand height wise.

Therefore, Desired result = Total ways of choosing models - (ways in which both fourth and sixth tallest are selected ) + (ways in which 4, 5, and sixth tallest are selected)
In the above, we add the (ways in which 4, 5, and sixth tallest are selected) because while subtracting the (ways in which both fourth and sixth tallest are selected ) , you would also have eliminated one favorable way. So we are compensating by adding it back.

Total ways of choosing models : $$7C5 = 21$$

Ways in which both fourth and sixth tallest are selected = select them both and select the additional 3 models from remaining 5: $$5C3 = 10$$

Ways in which 4, 5, and sixth tallest are selected = Select 4, 5, and 6 and select the additional 2 models from the remaining 4: $$4C2 = 6$$

So, desired ways : $$21 - 10 + 6 = 17$$
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Re: Out of seven models, all of different heights, five models [#permalink]

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24 May 2012, 21:05
7
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Hi,

Total number of ways 5 models can be selected = $$7C5$$ = 21
Total number of ways when 4th & 6th models stand together = 4C3 = 4 (4th & 6th are already chosen, 5th can't be chosen, so only 4 models are left from which 3 are chosen)

Total required ways = Ways of selecting all 5 models - no. of ways when 4th & 6th models are together,
=21 - 4 = 17

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Posts: 2
Re: Out of seven models, all of different heights, five models [#permalink]

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17 Nov 2012, 08:48
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

5 will be chosen to pose.
5 + 4 + 3 + 2 + 1 = 15

15 + 2 = 17

This is my 1st time doing Math again in 20 years after I got sober for 2 years
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GPA: 3.23
Re: Out of seven models, all of different heights, five models [#permalink]

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28 Dec 2012, 07:24
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Great question! Thank you!

My approach:

Case 1: 4 and 6 are selected...

This means 5 must also be selected. $$... 4 5 6 ...$$
Now, we need to select 2 more models from 4 remaining. $$=\frac{4!}{2!2!}=6$$
Since the models already have assigned heights then in each grouping possible, there is only 1 arrangement. $$=6 * 1 = 6$$

Case 2: 4 and NOT 6 is selected...
This means we need to select 4 from the remaining 5. $$=\frac{5!}{4!1!}=5*1=5$$

Case 3: 6 and NOT 4 is selected...
This means we need to select 4 from the remaining 5. $$=\frac{5!}{4!1!}=5*1=5$$

Case 4: 4 and 6 are not selected
$$=\frac{5!}{5!}=1$$

$$=6 + 5 + 5 + 1 = 17$$

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Director
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Re: Out of seven models, all of different heights, five models [#permalink]

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22 Apr 2013, 23:27
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

1. The size of the larger group is 7. So n is 7.
2. 5 models are to be selected from the larger group. So r is 5
3. The number of combinations without constraint is $$7C5 = 21.$$
4. Let us take the combinations that satisfy the opposite of the constraint i.e., models 4 and 6 being next to each other. This can happen in $$4C3$$ ways because 4 and 6 are always selected. Therefore n and r reduce by 2. But 5 is never selected. So n reduces by 1 more. Therefore, n is 4 and r is 3. And $$4C3$$ ways is $$4$$ ways.
5. (3) - (4) = $$21 -4 =17$$. This is the number of combinations with the constraint.

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Systematic Approaches

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Re: Out of seven models, all of different heights, five models [#permalink]

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23 Apr 2013, 00:42
riyazv2 wrote:
Total no. of ways "5 models can be selected from 7" = 7C5 = 21.

Fourth tallest and Sixth tallest can never stand adjacent to each other. Also, all selected models will be lined up from shortest to tallest.
Hence, we have to pull out the no. of ways 4 & 6 stand adjacent to each other from the above total, i.e = 21.

The following will be the way in which 4 & 6 stand close to each other.

1 2 3 4 6
1 2 4 6 7
1 3 4 6 7
2 3 4 6 7

So, the answer will be 21-4 = 17 ways.

Exactly how I solved it riyazv2.
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Re: Out of seven models, all of different heights, five models [#permalink]

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23 Apr 2013, 04:06
mbaiseasy wrote:
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Great question! Thank you!

My approach:

Case 1: 4 and 6 are selected...

This means 5 must also be selected. $$... 4 5 6 ...$$
Now, we need to select 2 more models from 4 remaining. $$=\frac{4!}{2!2!}=6$$
Since the models already have assigned heights then in each grouping possible, there is only 1 arrangement. $$=6 * 1 = 6$$

Case 2: 4 and NOT 6 is selected...
This means we need to select 4 from the remaining 5. $$=\frac{5!}{4!1!}=5*1=5$$

Case 3: 6 and NOT 4 is selected...
This means we need to select 4 from the remaining 5. $$=\frac{5!}{4!1!}=5*1=5$$

Case 4: 4 and 6 are not selected
$$=\frac{5!}{5!}=1$$

$$=6 + 5 + 5 + 1 = 17$$

Hi,

In case 2 , how a " Not 6 " is selected ? Will you please explain ?
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Re: Out of seven models, all of different heights, five models [#permalink]

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23 Apr 2013, 04:09
Bunuel wrote:
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Notice that since the fourth-tallest and sixth-tallest models cannot be adjacent and models are to stand in a line from shortest to tallest, then if we choose 4th and 6th we MUST also choose 5th to stand between them.

Also notice that since the models are to stand in a line from shortest to tallest then for any group of 5 models we choose, there will be only one arrangement possible: from shortest to tallest.

Hence we can have 3 cases:
Choosing 4th, 5th, and 6th tallest models and any 2 from 4 models left: $$C^3_3*C^2_4=6$$;
Choosing either 4th or 6th tallest models and any 4 from 5 models left: $$C^1_2*C^4_5=10$$;
Choosing neither 4th nor 6th tallest models: $$C^5_5=1$$;

Total: 6+10+1=17.

Excellent and clear approach.
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Re: Out of seven models, all of different heights, five models [#permalink]

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05 May 2013, 11:56
My Approach:

Let the models be 1,2,3,4,5,6 and 7.

1. If models 4,5 and 6 are included in the group of 5 models, then 2 models be selected from remaining 4 Models = 4C2 = 6
2. If model 4 is included and model 6 is not included in the group, then 4 models be selected from the remaining 5 Models = 5C4 = 5
3. If model 6 is included and model 4 is not included in the group, then 4 models be selected from the remaining 5 Models = 5C4 = 5
4. If model 4 and model 6 are not included in the group, then 5 models be selected from the remaining 5 Models = 5C5 = 1

Total = 17
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Re: Out of seven models, all of different heights, five models [#permalink]

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05 May 2013, 12:03
kabilank87 wrote:
mbaiseasy wrote:
eyunni 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 models cannot be adjacent, how many different arrangements of five models are possible?

A. 6
B. 11
C. 17
D. 72
E. 210

Great question! Thank you!

My approach:

Case 1: 4 and 6 are selected...

This means 5 must also be selected. $$... 4 5 6 ...$$
Now, we need to select 2 more models from 4 remaining. $$=\frac{4!}{2!2!}=6$$
Since the models already have assigned heights then in each grouping possible, there is only 1 arrangement. $$=6 * 1 = 6$$

Case 2: 4 and NOT 6 is selected...
This means we need to select 4 from the remaining 5. $$=\frac{5!}{4!1!}=5*1=5$$

Case 3: 6 and NOT 4 is selected...
This means we need to select 4 from the remaining 5. $$=\frac{5!}{4!1!}=5*1=5$$

Case 4: 4 and 6 are not selected
$$=\frac{5!}{5!}=1$$

$$=6 + 5 + 5 + 1 = 17$$

Hi,

In case 2 , how a " Not 6 " is selected ? Will you please explain ?

Case 2: If 6 is not included in the group then 4 and 6 cannot stand adjacent .
Case 3: if 6 is not included in the group then 4 and 6 cannot stand adjacent.
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Re: Out of seven models, all of different heights, five models [#permalink]

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08 Sep 2014, 03:11
Another approach:

One case would be when the fifth tallest has to be there = 6C4 ways of selection possible= 15

Second case would be ,if the fifth tallest is not there, 4th tallest is also not there= 5C5 ways = 1

Third case would be if the fifth tallest is not there, sixth tallest is also not there= 5C5 ways = 1

Total number of ways =17.
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Systematic Approaches

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Re: Out of seven models, all of different heights, five models [#permalink]

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03 Mar 2018, 13:42
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Re: Out of seven models, all of different heights, five models   [#permalink] 03 Mar 2018, 13:42
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