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Comment on Handling Restrictions
Why do you assume that
The key word here is "arrange
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.
Hi,
I sometimes get a bit confused between mnp rule & !factorial principle. Can you please throw light on the similarities and distinction between them?
Can you please elaborate. I'm
Can you please elaborate. I'm not sure what you mean by the "mnp rule" and the "factorial principle."
Fundamental counting
We can use either approach,
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
The key word here is "arrange
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
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?
Yes, in that case the word
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
Happy to help!
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...
Cheers,
Brent
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
Hi ipshitasaha,
Hi ipshitasaha,
You'll find my solution here: http://www.beatthegmat.com/six-children-a-b-c-d-e-and-f-t279370.html
Cheers,
Brent
Can you please provide an
My solution
x = 3
x-1
y-10
z-9
=90
end with 3 (z=3)
x-4
y-10
z-1
= 40
both
x-1
y-10
z-1
= 10
sum: 140
Question link: https:/
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.
Cheers,
Brent
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?
https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begin-with-a-3-end-with-a-268515.html
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
Question link: https:/
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
Cheers,
Brent
https://gmatclub.com/forum
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?
Question link: https:/
Question link: https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begi...
Perfect approach!
Cheers,
Brent
Can we do FCP with
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
Yes, we use the Fundamental
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.
Cheers,
Brent
Hi Brent. Good Evening.
In question - https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begin-with-a-3-end-with-a-268515.html,
I understood the explanation you have given but wondering if it would be a question for a 5 digit number or more then the similar solution would have consumed more time, so I was thinking if you can solve it using stages and applying fundamental counting principle. I hope that's possible, can you please solve it with this approach?
Question link: https:/
Question link: https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begi...
I use the fundamental counting principle in my solution (https://gmatclub.com/forum/how-many-positive-integers-less-than-500-begi...). I just used the strategy to count the outcomes that DON'T satisfy the restriction, and then I subtracted that amount from the total number of integers less than 500.
We can use the same strategy to answer the question a different way, but we still need to consider different cases:
1-digit numbers that begin with a 3, end with a 3, or both: 3 (1 number)
2-digit numbers in the form 3X
2-digit numbers in the form X3
2-digit numbers in the form 33 (1 number)
3-digit numbers in the form 3XX
3-digit numbers in the form XX3
3-digit numbers in the form 3X#
Hello Brent,
In advance thank you for your time. Can you please help me with this question from GMAT Official Practice :
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
I'm happy to help.
I'm happy to help.
Here's my full solution: https://gmatclub.com/forum/if-a-committee-of-3-people-is-to-be-selected-...
I like the first one!