Lesson: Handling Restrictions

Comment on Handling Restrictions

Why do you assume that letters cannot be repeated
gmat-admin's picture

The key word here is "arrange." So, we're taking 5 things (letters) and determining the number of ways to move (arrange) those 5 letters around.

I sometimes get a bit confused between mnp rule & !factorial principle. Can you please throw light on the similarities and distinction between them?
gmat-admin's picture

Can you please elaborate. I'm not sure what you mean by the "mnp rule" and the "factorial principle."

Fundamental counting principle vs arranging n unique objects
gmat-admin's picture

We can use either approach, since they both yield the same results.

With the Fundamental Counting Principle, we can see WHY n unique objects can be arranged in n! ways.

Or we can just use the formula (n unique objects can be arranged in n! ways)

why 5*5*4*5*5 is wrong because nothing is mentioned about repetition of words.
gmat-admin's picture

The key word here is "arrange." So, we're taking 5 things (letters) and determining the number of ways to move (arrange) those 5 letters around (with a restriction).

So arrange implies no repetition in that case?

Request you to also explain why have we ruled out repetition in the following question: If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of four-digit extensions that the company can have?
gmat-admin's picture

Yes, in that case the word "arrange" implies no repetition.

In your second question, the keyword is "ALL" as in, "...each of the extensions uses ALL four of the digits 1, 2, 3, and 6..."

So, the digits 1, 2, 3 and 6 must ALL be used in the four-digit extension. This means there cannot be any repetition.

Hi Brent,

Can you pls help me understand the solution to the following question, using the above formula:

Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

(A) 240
(B) 480
(C) 720
(D) 1440
(E) 3600

Thanks, Ipshita
gmat-admin's picture

Happy to help!

I have provided a step-by-step solution here: https://gmatclub.com/forum/seven-children-a-b-c-d-e-f-and-g-are-going-to...


Hi Brent,

Pls help me understand the solution to this question using the technique explained in the video:

Six children, A, B, C, D, E, and F, are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

Thanks, Ipshita
gmat-admin's picture

Hi ipshitasaha,

You'll find my solution here: http://www.beatthegmat.com/six-children-a-b-c-d-e-and-f-t279370.html


Can you please provide an answer to this question I got 140, the answer says its 149.

My solution
x = 3
end with 3 (z=3)
= 40
= 10
sum: 140
gmat-admin's picture

Question link: https://greprepclub.com/forum/how-many-positive-integers-less-than-500-b...

The question: How many positive integers less than 500 begin with a 3, end with a 3, or both?

I don't entirely follow your solution, but it appears that you are considering only 3-digit numbers.
So, for example, you don't include 35 as a possible value.
Likewise, the single-digit number 3 meets the condition.


Hi Brent,

I was trying to solve the below question using the counting strategy but I am not getting the right answer Her's my approach, could you help me identify what am I missing here?

1 digit no with 3: 1
2 digits no with 3: (_3 = 8 ways) + (3_ = 10 ways) = 18 ways
3 digits no with 3:120 ways
A: 3 _ _ = 9*9 = 81 ways
B: 3 3 _ = 10 ways
C: 3 _ 3 = 9 ways
D: _ _ 3 = 2*10 = 20 ways
Total = 139 ways
gmat-admin's picture

Question link: https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begi...

You almost had it!!
The mistake is at the very end, where you write: D: _ _ 3 = 2*10 = 20 ways
The hundreds digit can be 1, 2 or 4.
So, the last part should read as follows:
D: _ _ 3 = 3*10 = 30 ways


did this in a little different way
single digit number = 1
two digit number starting with other than 3 but ending with 3 = 1,2,4,5,6,7,8,9 8*1= 8
two digit number starting with 3 = 1*10= 10
3 digit number starting with 3 but last digit not 3 = 1*10*9= 90
3 digit number ending with 3 = 3*10*1 = 30
3 digit number ending and starting with 3 =1*10*1 = 10
total ways = 10 +30 +90 +10 +9 = 149
sir is this approach correct?
gmat-admin's picture

Can we do FCP with restrictions?

Numbers beginning with 3:
1 X 10 x 9 (since it cannot end with 3)

Numbers ending with 3: (begins with 4,2,1,0)
4 x 10 x 1

Numbers with 3 both beginning and end:
1 x 10 x 1

add them up = 99 + 40 + 10 = 149
gmat-admin's picture

Yes, we use the Fundamental Counting Principle with restrictions.
Your solution (to the question https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begi...) is perfect.


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