A lift with 7 people stops at 10 floors. People varying from zero to

I remember there was a formula for this during my preparation for CAT (Indian Management Exam) I believe it was (n+r-1) c (m-1)

Here 7 people for 10 floors so (10+7-1) = 16

m = (0 to 7) ~ 8 = m-1 = 7

so 16c7 should be B

looking for OE and not just an educated guess

Looking at the question from a different perspective:
How many way can we distribute 7 people to 10 floors?

To solve this question we can use a method know as STARS AND BARS.
This method is used, more in general, to identify the number of ways to arrange n indistinguishable items into k distinguishable bins. In our case n=7, k=10.

Let's forget for a moment about people and floors and rephrase the question into:

7 stars (or items)
9 bars (or separators) 9 = 10-1 --> imagine you have 10 bins, each divided from the next by a separator, therefore 10-1 = 9 separators (bars)

Example
*| * | ** | | | | | | |*** --> here we have 1 item in bin #1, 1 item in bin #2, 2 items in bin #3 and 3 items in bin #10 (0 item for all the others bins)
******* | | | | | | | | | --> here we have 7 item in bin #1, no item in each of the other bins

The method uses the following formula to solve the possible ways to distribute the n identical items into k different bins:
(n+k-1)C(k-1)

In our case since n = 7; k = 10
we get (7+10-1)C(10-1) --> (16)C(9), or equivalently 16C7

Coming back to stars&bars method, the same formula can be applied using directly stars (items) and bars (separators).
Therefore as follows:
(stars+bars)C(bars) --> (16)C(9) or equivalently 16C7

Answer is B. --> 16C7

Given: A lift with 7 people stops at 10 floors. People varying from zero to seven go out at each floor.

Asked: The number of ways in which the lift can get emptied, assuming each way only differs by the number of people leaving at each floor, is

The question is similar to finding number of solution to the equation : -
x1 + x2 + x3 + x4 + .... + x10 = 7

The number of ways in which the lift can get emptied = (10 + 7 - 1)C7 = 16C7

IMO B

We need to distribute 7 identical people (since each way differs only in the number of people leaving at each floor and not which people are leaving so all people are assumed identical) into 10 distinct groups (10 distinct floors).

To distribute identical things into distinct groups, we use the method of partitions. We need 9 partitions to make 10 distinct groups.
So we have the arrange 7 people and 9 partitions.
PPPPPPP|||||||||

We can do that in 16!/7!*9! ways which is same as 16C7

Answer (B)

Check out the concept of distribution of distinct/identical objects into distinct/identical groups in my Combinatorics module.

Hi,
How is 16c9=16c7?
someone please respond

\(16C9 = \frac{16!}{(16 - 9)!*9! }= \frac{16!}{7!*9! }\)

\(16C7 = \frac{16!}{(16 - 7)!*7! }= \frac{16!}{7!*9! }\)

Generally, \(nC_r =nC_{n-r}\).

for distributing 'n' identical amongst r , such that any person might get any no of things including 0
so here n is people and r is floors
(n+r-1) C (r-1)
here n is 7 and r is 10
16c9
Bunuel I think answer option should have 16c9 as an option instead of 16c7 ;
though value of 16c9 = 16c7....

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.