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## Counting problem

tagged by: rolandprowess

This topic has 5 member replies
rajs.kumar Senior | Next Rank: 100 Posts
Joined
09 Jun 2006
Posted:
87 messages
2

#### Counting problem

Wed Oct 25, 2006 3:41 am
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?

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

### Top Member

Roland2rule Legendary Member
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Sun Nov 05, 2017 12:17 am
we are to choose five models from a group of seven models of different heights. if the fourth-tallest and sixth-tallest model cannot be adjacent.
Now, total number of selections possible if any model can be with any model is
$$7C_5=\frac{7!}{\left(7-5\right)!5!}=\frac{7!}{2!5!}=\frac{\left(7\cdot6\right)}{2!}=21ways$$
i.e in total, there are 21 possible selections. Now, the only conditions when the fourth-tallest and sixth-tallest model CAN be adjacent is
(1) when both of them are selected together
(2) when the fifth-tallest model is not chosen

This takes some reasoning because if both of them are chosen and the fifth-tallest model is also chosen, then they can never be adjacent.
is this explanation logical to you? If YES, .... then let's proceed
let n be the number of ways by which the fifth-tallest model is not chosen, and the fourth-tallest AND the sixth-tallest are chosen. In this case, we will be left with four models to choose three models from , as two positions have been occupied by the fourth-tallest and sixth-tallest, and we have excluded the fifth-tallest model from the selection.
$$n=4C_3=\frac{4!}{\left(4-3\right)!3!}=\frac{4\cdot3!}{1!3!}=4$$
if the two of them are to be adjacent, the number of selections will be 4.
Therefore, the number of selections in which the two of them cannot be adjacent is
=total number of selection possible - n
=21-4
=17 possible selections

Badri Junior | Next Rank: 30 Posts
Joined
09 Mar 2007
Posted:
14 messages
1
Fri Mar 09, 2007 7:22 pm
the assumption :

"...the only way is if they are arranged in the 3rd and 4th position ..."

is not quiet right.

Barring '5' we have '1', '2', '3', '4', '6', '7' and we need to select group of 5 out of these 6 models. Now Since '4' and '6' must be selected to violate the rule; so, we need to choose 3 out of 4 i.e. 3 models out of '1', '2', '3', '7'
which gives 4C3 = 4 ways.
So, the right combination which will violate the rule is:
(increasing order of height and where '4' and '6' are adjacent to each other)

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

i.e. only 4 groups will violate the rule.

now total number of ways of making 5 groups out of 7 people :
7C5 = 21

but out of these 21 groups 4 groups will not meet the requirement hence total number of groups which will follow the rule:

= 21 - 4
= 17

ajith Legendary Member
Joined
21 Sep 2006
Posted:
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Test Date:
April 2010
Target GMAT Score:
740
Wed Oct 25, 2006 5:35 am
Case 1:

4th tallest and 6 th tallest models are selected. 5 th highest model has to be selected in this case

Now this is a question of selecting 2 out of remaining 4 models

this can be done in 4c2=6 ways

Case 2:

Strictly One of 4th tallest and 6th tallest model is selected. We can select other 4 members from remaining 5. This will meet all the conditions in the question.

this can be done in 2c1*5c4=10 ways ..

Case 3

Both 4th and 6th tallest persons are not selected

this can be done in 5c5=1 way

Total no: of ways

10+6+1

=17

rajs.kumar Senior | Next Rank: 100 Posts
Joined
09 Jun 2006
Posted:
87 messages
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Fri Oct 27, 2006 1:52 am

another approach

# of way in which 4 and 6 dont stand together = total number of ways in which 5 out of 7 models can be selected - # of ways in which 4 and 6 can stand together

total number of ways = 7c5 = 7c2 = 21

4 and 6 together => 4 and 6 selected and 5 not selected.

64 - - -

Remaining 3 positions can be filled in 4c3 = 4c1 = 4 ways (choices available 1237 and number of position available = 3)

Answer = 21 - 4 = 17

anuroopa Junior | Next Rank: 30 Posts
Joined
30 Sep 2006
Posted:
16 messages
1
Fri Mar 09, 2007 1:45 pm
here's my doubt:
if the descending order in terms of height was represented as 1<2<3<4<5<6
if 4 and 6 have to come together - the only way is if they are arranged in the 3rd and 4th position - which then means 7 must follow

having said that - there are only 3 ways to fill the positions in front of 4 - as 5 would mean violating the tallest to shortest constraint

12467
13467
23467

I cant think of the 4th way - can somebody help me sort this

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