lets have a discussion

[sana.noor]

i have two questions here, both are same in the nature but the solution of the two are different. both questions are solved by "separator" method and i have issue with the formula.
1) 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is
a. 30 b. 210 c. 15 d. 35 e. 40
after distributing 12 oranges evenly we are left with 4 oranges. Number of grouping simply becomes number of ways to arrange 4 oranges and 3 separators between themselves = (4 + 3)!/(4!*3!) = 35

2) How many of the four-digit numbers with non-zero digits have the sum of their digits as 12?
165
330
132
440
170
This can be done in (11P3)/3! = (11!)/(8! * 3!) = (11*10*9)/6 = 11*15 = 165 ways (WHY IT IS 11P3 WHY NOT 12P3)

my question is that the first question is using total number of sticks and separator as 4+3!, i understand this, but why the second question is using one less separator?

[theCodeToGMAT]

it cannot be 12!/(9!*3!) because:

- you are NOT considering the main condition .. i.e. "the sum should be 12"
- 9! would mean that all 9 digits are same.. for example Q1.. you are considering oranges.. so you did 4!

I don't know how to solve this question..but trying to find solution

[GMATGuruNY]

2) How many of the four-digit numbers with non-zero digits have the sum of their digits as 12?
165
330
132
440
170

why the second question is using one less separator?

A sum of 12 units is to be DISTRIBUTED among 4 digits.
None of the digits can be greater than 9.
This is the equivalent of the following problem:

How many ways can 12 identical chocolates be distributed among 4 children A, B, C and D if no child can receive more than 9 chocolates?

To guarantee that no child receives more than 9 chocolates, first distribute 1 chocolate to each child.
Now count the number of ways to distribute the REMAINING 8 CHOCOLATES to the 4 children.

The following is called the SEPARATOR method.

8 identical chocolates are to be separated into -- at most -- 4 groupings.
Thus, we need 8 chocolates and 3 separators:
OO|OO|OO|OO

Each arrangement of the elements above represents one way to distribute the 8 chocolates among 4 children A, B, C and D:
OO|OO|OO|OO = A gets 2 chocolates, B gets 2 chocolates, C gets 2 chocolates, D gets 2 chocolates.
OOOOOO|||OO = A gets 6 chocolates, D gets 2 chocolates.
OOOOOOOO||| = A gets all 8 chocolates.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 11 elements above (the 8 identical chocolates and the 3 identical separators).
The number of ways to arrange 11 elements = 11!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 8! (the number of ways to arrange the 8 identical chocolates) and by 3! (the number of ways to arrange the 3 identical separators):
11!/(8!3!) = 165.

The correct answer is A.

For two similar problems, check here:

https://www.beatthegmat.com/algebra-t215423.html#622071

[theCodeToGMAT]

Superb Mitch!!!

[sana.noor]

thanks mitch!

[sana.noor]

thanks mitch!