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A researcher plans to identify each participant in a certain

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Re: A researcher plans to identify each participant in a certain [#permalink]

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New post 19 Dec 2017, 19:36
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dave13 wrote:
Bunuel wrote:
kevn1115 wrote:
Hi Bunuel,

I'm confused on when you show that n! = (n-2)!*(n-1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n-2)!?

Thanks.!


\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.


n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n-4)*(n-3)(n-2)(n-1)n. To simplify \(\frac{n!}{2!(n-2)!}\) I wrote n! as (n-2)!*(n-1)*n this enables us to reduce by (n-2)! to get \(\frac{(n-1)n}{2}\).

Hope it's clear.


Bunuel -Thank you ! Now when I`ve reviewed the whole thread and still trying to understand some moments - why do you write it in this order 1*2*...*(n-4)*(n-3)(n-2)(n-1)n and not vice versa like this 1*2*...n(n-1)(n-2)(n-3)(n-4) etc ...also why you say" notice that n!=(n-2)(n-1)n" yes it as n! is in numerator as per formula and unlike formula, you simplify n! = n!/2!(n-2)! into != (n-2)(n-1)n" / 2!(n-2)!
first cant not "notice" the important detail you are trying to imply by pointing at this ---> n!=(n-2)(n-1)n - can I be helped with further explanation to understand this "notice" because in other combination formulas we didn't apply such simplification :? :)


1. n! is the product of integers from 1 to n, inclusive. So, n! = 1*2*...*(n-4)(n-3)(n-2)(n-1)n (1 is the smallest and n is the largest). Yes, you can write this in any order but it does not change anything.

2. n! = (n-2)!*(n-1)*n because (n-2)! = 1*2*...*(n-4)(n-3)(n-2), so (n-2)!*(n-1)*n = [1*2*...*(n-4)(n-3)(n-2)](n-1)n = n!

3. We can write \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\) whenever it's necessary.

Hope it's clear now.
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A researcher plans to identify each participant in a certain [#permalink]

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New post 09 Jan 2018, 07:12
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


We can use just logic to solve this question.

Here we are given that all letters and we need to get the least number of letters for 12 participants and all the letter should in alphabetical order.

Let's choose 4 letters since this is the least number.
A B C D
Then we can choose two letters again , AB, AC, AD, BC, BD, BD - total 6 ways. ( Note here we can't BA since we need to follow alphabetical order )
Total 4 + 6 = 10 and this suits only for 10 participants.

Let's choose 5 letters.
A B C D E - 5 ways
AB AC AD AE BC BD BE CD CE DE - 10 ways . We got 15 ways and 15 different codes are sufficient to give the code for 12 participants.

Hence 5.
A researcher plans to identify each participant in a certain   [#permalink] 09 Jan 2018, 07:12

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