First each child gets at least one gift, so the first 6 gifts are distributed to 6 children (1 each) - 1 way

Now you have 4 gifts remaining and 6 people who can take these gifts.

xxxx |||||

x be the gifts and | represent people (6 people so 5 slots), they can be arranged in 9!/5!4! ways (or) 9C5 ways


The direct formula is the one you mentioned, pretty straightforward, you get 9C5

@ashwin
Shouldn't the formula be (n-1)C(r-1) in this case? I mean, if we do not consider that each child has first been given 1 gift each at the beginning

The formula you mentioned is ONLY applicable when 'n' things are to be distributed to 'r' people without any restrictions, i.e 1 person can get all 'n' and others get 0 and so on.

We are given a condition that each child receives at least 1 gift here, so the formula cannot be directly applied in this case.

Always make sure you apply formulas correctly when using them, IMO a formulaic approach to P&C often gives misleading answers for the test makers frame questions such as this, where you could go wrong when you use the formula directly. A more logical approach always gets you close to the answer, even for a educated guess.

Once we recognize the problem tests us on (n+r-1)C(r-1), we should ensure we the formula is applicable to our scenario. Hence we initially distribute the first 6 gifts and then use the formula.

A more general way of approaching P&C problems is to always account for *any given conditions* in the problem before using generic counting methods/slot method or even for formulas.

kakz wrote:@ashwin
Shouldn't the formula be (n-1)C(r-1) in this case? I mean, if we do not consider that each child has first been given 1 gift each at the beginning

@Ashwin
I think u slightly misunderstood the question. Kakz said the formula is (n-1)C(r-1) and not
(n+r-1)C(r-1). Try this problem by doing (n-1)C(r-1) which is the formula for when each person gets
atleast 1 and the remaining can be distributed in any way. You'll get the answer 9C5

(n+r-1)C(r-1) is, like u said, the formula which includes distributions where one person gets all etc. Basically, it includes all the restricted possibilities.

Try using the (n-1)C(r-1) on the following problem. U'll see what I'm talking about

https://www.beatthegmat.com/combination-t41362.html



P.S. I've seen a quite similar funda on the Arun Sharma book on CAT Quant. Haven't u seen this funda there?

Hey sorry guys... I wasn't really aware of that formula. Guess its applicable for this problem here.

Somehow never had a formulaic approach for P&C, and at least for the GMAT you don't need to know complex P&C rules.

In my opinion this is quite an advanced problem and you would see it unless you're scoring high in quants.