BTGModeratorVI wrote: ↑

Sun Jul 19, 2020 1:33 pm

5 people including A and B line up in a row. How many possible cases are there such that at least one person stands between A and B?

A. 24
B. 36
C. 48
D. 60
E. 72

Answer: E
Source: Math Revolution

Without any constraints, 5 people can line up in 5! = 120 ways.

Let's take AB together 4; thus, there are 4 people. Four people can line up in 4! = 24 ways. But AB and also line up as BA, so, the total no. of such ways = 2*24 = 48

Required # of ways = 120 – 48 = 72

Correct answer: E

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Sun Jul 19, 2020 1:33 pm

5 people including A and B line up in a row. How many possible cases are there such that at least one person stands between A and B?

A. 24
B. 36
C. 48
D. 60
E. 72

Answer: E
Source: Math Revolution

Here's a different approach:
Let's let the 5 people be A, B, C, D and E

Take the task of arranging the 5 people and break it into stages.

Stage 1: Arrange C, D and E in a row
We can arrange n items in n! ways.
So, we can arrange 3 people in 3! (aka 6) ways.
We can complete stage 1 in 6 ways

IMPORTANT: For each of the 6 arrangements of C, D and E, we can place a space on either side of each person.
For example, for the arrangement DEC, we can add spaces as follows: _D_E_C_
We can now place A in one of these spaces, and place B in one of these spaces.
This will ensure that A and B do NOT sit together

Stage 2: Place person A in one of the spaces.
There are 4 spaces to choose from. So, we can complete stage 2 in 4 ways

Stage 3: Place person B in one of the spaces.
There are 3 spaces remaining.So, we can complete stage 3 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange all 5 people) in (6)(4)(3) ways (= 72 ways)

Answer: E