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

Here is how I think...

7 models say 1,2,3,4,5,6,7

Now find out the arrangements where 4 and 6 are NOT adjacent or seat together = total arrangements - Seat together

total arrangements = 7C5 = 21
Seat together = If assume that 4 and 6 are selected in the pool already, need to find out other three members. Therefore we have 7-3 = 4 members left. I deduct 3 because we have to neglect 5 also otherwise 4 and 6 can't be adjacent (the five models are to stand in a line from shortest to tallest - here is the significance). Now select 3 people from 4 members in 4C3 ways.
Lets form the equation:
the arrangements where 4 and 6 are NOT adjacent or stand together = 7C5 - 4C3 = 21 - 4 = 17.

When we choose 4 and 6 and they are not adjacent means that we must choose 5 too (to stand between them). So for this case we must choose 4, 5, and 6 (3C3) and 2 other from 4 left (4C2) = 3C3*4C2=4C2.

This can be solved in another way:

If we choose 4 and 6, we must also choose 5 (to stand between them) =3C3*4C2=4C2=6
We can choose either 4 or 6 = 2*1C1*5C4=10
We can choose neither 4 nor 6 = 5C5=1

6+10+1=17.

Answer: C (17).

Hope it helps.

I got 17, is it correct?

Ways in which you can sit all models in oder: 7!/5!2! = 21
Then we figure out in how many of those 21 options model 4 and 6 are sitting together.

I usually draw something like this:
Option 1: _ _ _ _ _ Model 4, 6 could take this spaces, leaving last spot for model 7. You can have 3 in which you can arrange the first 3 models in the first 2 spots = 3

Option 2: _ _ _ _ _ Model 4, 6 could also be on the last 2 spots, but there's only one option on this one, because there are only 3 models to fill the first 3 spots = 1

21 - (3+1) = 17

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

Answer (c)

lets say 1 2 3 4 5 6 7 are models with 1 to 7 from smallest to tallest.

no of ways to slect any 5 models from 7 models =7C5 =21

no of arrangement for any selection would be =1

there for no of arrangement for 21 selection =21

now 4 and 6 should not be adjusant ,so lets subtract the cases in which 4 and 6 are adjusant.

4 and 6 can be adjusant only when 5 is not selected . so we have decided about selecting two models 4 & 6
and not selecting 5 .
no of models left (1 2 3 4 5 6 7)- (4,6 )-5 =1 2 3 7 only four models are left for selection
so to make a group of 5 models we have to select 3 out of 4 (or drop 1 out of 4) =4C3 (or 4C1)=4
21-4=17

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


Soln:
Total number of ways of choosing 5 out of 7 models is = 7C5
Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3
Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2


= 7C5 - 5C3 + 4C2
= 21 - 10 + 6
= 17

Ans is B

Brilliant, had the same idea, but didn't know how to model it with combinations.

Since the models are arranged according to their heights and since no two models have the same height, as soon as the five models are chosen, the arrangement of the photo is determined. Thus, without other restrictions, there are 7C5 = 7!/(5!*2!) = (7 x 6)/2 = 21 arrangements possible.

Let’s determine in how many of these 21 restrictions do the 4th tallest and the 6th tallest models stand together. Notice that in order for the 4th tallest and the 6th tallest model to stand together, the selection must include the 4th tallest and the 6th tallest model, but must not include the 5th tallest model. Assuming the 4th tallest and 6th tallest models are already chosen, we must choose 3 additional models from the remaining 4 models (all models besides the 4th, 5th and 6th tallest models). The number of ways we can do this is 4C3 = 4.

Thus, in 21 - 4 = 17 of the arrangements, the 4th tallest and 6th tallest models do not stand together.

Answer: C

Could someone explain me the part highlighted in red? Thanks

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