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# A certain club has 20 members. What is the ratio of the memb

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Manager
Joined: 26 Mar 2017
Posts: 115
Re: A certain club has 20 members. What is the ratio of the memb  [#permalink]

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15 Jun 2017, 10:00
Bunuel wrote:
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?

$$\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}$$

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is $$C^2_3=3$$.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.

hey I got the correct answer but was just wondering why the members have to be different, i.e. why we are using combinations ?

What am I missing here ?
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A certain club has 20 members. What is the ratio of the memb  [#permalink]

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22 Jun 2018, 13:57
So the formula that we have to remember when it comes to combinatorics in which case sequence does not matter is this:

n!
---------
(k!)((n-k)!)

in which n is the total number of units and k is the number of units per set.

--------------------------------------------------------------------------

So in the case of our problem, n = 20 and for the first set k = 5. In the second set k = 4. So putting it all together, it looks like this:

20!
---------
5!((20-5)!)
--------------
20!
---------
4!((20-4)!)

Simplify:

20!
---------
5!((15!)
-------------
20!
---------
4!((16!)

Simplify:

20! 4!(16!)
------- x -------
5!(15!) 20!

Simplify:

4!(16!)
-------
5!(15!)

Solve:

4 * 3 * 2 * 1 * 16 * 15! 16
-------------------- --> ---- --> or 16 : 5
5 * 4 * 3 * 2 * 1 * 15! 5

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Re: A certain club has 20 members. What is the ratio of the memb  [#permalink]

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28 Feb 2019, 07:08
Top Contributor
milind1979 wrote:
A certain club has 20 members. What is the ratio of the member of 5-member committees that can be formed from the members of the club to the number of 4-member committees that can be formed from the members of the club?

A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4

5-member committees
Since the order in which we select the committee members does not matter, we can use COMBINATIONS
We can select 5 people from 20 people in 20C5 ways
20C5 = (20)(19)(18)(17)(16)/(5)(4)(3)(2)(1)

4-member committees
Since the order in which we select the committee members does not matter, we can use COMBINATIONS
We can select 4 people from 20 people in 20C4 ways
20C4 = (20)(19)(18)(17)/(4)(3)(2)(1)

The RATIO = (20)(19)(18)(17)(16)/(5)(4)(3)(2)(1)]/[(20)(19)(18)(17)/(4)(3)(2)(1)]
= (20)(19)(18)(17)(16)/(5)(4)(3)(2)(1)][(4)(3)(2)(1)/(20)(19)(18)(17)]
=(20)(19)(18)(17)(16)(4)(3)(2)(1)/(20)(19)(18)(17)(5)(4)(3)(2)(1)
=(20)(19)(18)(17)(16)(4)(3)(2)(1)/(20)(19)(18)(17)(5)(4)(3)(2)(1)
=16/5

Cheers,
Brent

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Re: A certain club has 20 members. What is the ratio of the memb  [#permalink]

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04 Mar 2019, 20:18
milind1979 wrote:
A certain club has 20 members. What is the ratio of the member of 5-member committees that can be formed from the members of the club to the number of 4-member committees that can be formed from the members of the club?

A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4

The ratio is the following:

(20C5)/(20C4) = [(20 x 19 x 18 x 17 x 16)/5!]/[(20 x 19 x 18 x 17)/4!]

= [(20 x 19 x 18 x 17 x 16)/5!] x [4!/(20 x 19 x 18 x 17)]

= 16/5

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Re: A certain club has 20 members. What is the ratio of the memb   [#permalink] 04 Mar 2019, 20:18

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