If each participant of a chess tournament plays exactly one

We are basically told that we can choose 153 groups of two players out of n players, thus \(C^2_n=153\) --> \(\frac{n!}{(n-2)!2!}=153\) --> \(\frac{(n-1)n}{2}=153\) --> \((n-1)n=306\) --> \(n=18\).

Answer: D.

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If the number of participants is 3 (say A, B, C) the number of games played will be 2 (A plays against B and C) + 1 (B plays against C) = 3
Using the same logic, if the number of participants is n, the number of games played will be (n-1) + (n - 2) + (n - 3) + ... 3 + 2 + 1

Given that this sum = 153 = 1 + 2 + 3 + ... ( n - 1)

Sum of first m positive integers is given by m(m+1)/2. So sum of first (n-1) positive integers is (n-1)*n/2

153 = (n-1)*n/2
(n-1)*n = 306
17*18 = 306 (We know that 15^2 = 225 so the two consecutive numbers must be greater than 15. Also, 20^2 = 400 so the two numbers must be less than 20. The pair of numbers in between 15 and 20 whose product ends with 6 is 17 and 18)

So n = 18

Answer (D)

Your method is absolutely fine. Your logic of multiplying n players by (n-1) games and dividing by 2 is great. The issue with your approach is that you need to calculate for every option. If the numbers were a little larger, you would end up doing calculations a number of times.
For each option, you are doing this calculation (n-1)*n/2
(A) 14*15/2
(B) 15*16/2
etc
You are trying to find the option that will give you the result 153. I have done the same thing. I get (n-1)*n/2 = 153.
Instead of trying all options, you should multiply 153 by 2 to get 306 and then see which two numbers will end in 6. That way you will save a lot of calculations.

14*15 - No
15*16 - No
16*17 - No
17*18 - Yes
18*19 - No

As for the approach used by me to arrive at n*(n-1)/2:
Think of it this way - you make all participants stand in a straight line. The first one comes up and play a game with everyone else i.e. (n-1) games and goes away. The next one comes up and plays a game with all remaining people i.e. (n-2) people and goes away too. This goes on till last two people are left and one comes up, play a game against the other and they both go away. Hence, they end up playing
(n-1) + (n - 2) + .... 3 + 2 + 1 games (total number of games)

You must learn that sum of first m positive integers is given by m(m+1)/2 (very useful to know this)
We need to sum first (n-1) numbers so their sum will be (n-1)n/2

Bunuel has used the combinatorics approach to arrive at n(n-1)/2. There are n people and you want to select as many distinct two people teams as you can (since each person can play against the other only once)
In how many ways can you do that? nC2 ways
nC2 = n(n-1)/2

So the calculation involved is the same in every case. The method of arriving at the equation is different.

Thanks Karishma

I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices long-hand.

If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games:
14*15 = 210
201/2 = 105, not 153, next

If B, 15*16 = 240
240/2 = 120, not 153, next

If C, 16*17 = 274
274/2 = 137, not 153, next

If D, 17*18 = 306
306/2 = 153 ANSWER D

Am I missing any important concepts by answering the question with this long-hand method as opposed to your more structured approach?

Kind regards

Derek

\(\frac{P!}{2!(P-2)!}=153\)
\(\frac{P * (P-1) * (P-2)!}{(P-2)!}=2*9*17\)
\(P*(P-1) = 18*17\)
\(P = 18\)

Answer: D

both are actually doing the same thing. You approach is a logical one while Karishma's is a structured one. But eventually both are doing same thing:
n*(n-1)/2 = 153 ( or 17*18/2 =153)

Num of games - 153, The total number of players that can be chosen as a pair would be nC2.

Now look through the options & start form the middle number as the options are in ascending order - 17C2 = 17*16/2 = 6 as last digit (17 *8) ignore
Next number 18c2 = 18*17/2 = 3 as last digit this is the answer

Choice D

x!/2!*(x-2)!=153, so

Backsolving looks better:

get C=17
17!/2!*15!=17*16/2=136

go D=18
18!/2!*16!=18*17/2=153 (correct)

D

bunuel how did you simply this equation?

SandhyAvinash , maybe this post will help.

Here are a couple of sites I found that discuss the simplification of factorials with variables. . .

This site is thorough and easy to understand:

https://chilimath.com/lessons/intermediate-algebra/simplifying-factorials-with-variables/

This next site's treatment is shorter than that above, but I have seen others link to this site. (Other experts may have linked to chilimath as well. I ran the math as it was explained on chilimath. The math is fine.)

https://www.purplemath.com/modules/factorial.htm

Hope they help!

P.S. I think questions here work better if you type the symbol @ before the username in your post. The name will show up in blue, like this SandhyAvinash.

If the person follows topics via email, s/he will get notified. S/he will not get notified with just SandhyAvinash, for example.
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It's a 10 second problem once you understand that the question is simply asking if nC2=153 then what is the value of n?

consider only 2 players; A and B
then no of matches "If each participant of a chess tournament plays exactly one game with each of the remaining participants"=1 (A vs B)
as any particular match involves 2 players here..because each player plays against each of the remaining participants separately...
so we only have to select 2 players...a simple combination.. to count a possible match from amongst the available players...

If there are 2 players only...then selecting 2 out of 2=2C2=1 (AB)
if there are 3 players only....then selecting 2 out of 3=3C2=3 (AB,AC,BC)
if there are 4 players only....then selecting 2 out of 4=4C2=6 (AB,AC,AD,BC,BD,CD)
Similarly...
if there are n players only....then selecting 2 out of n=nC2=n!/2!(n-2)!

here..nC2=n!/2!(n-2)!=153...put values back from options..we get n=18

Further....had a match involved 3 players rather than 2....then the no. of matches would have been nC3...
i.e., selecting 3 out of n players to play a match.

Similarly....had a match involved 4 players rather than 2....then the no. of matches would have been nC4...
i.e., selecting 4 out of n players to play a match.

Thus,
had a match involved r players rather than 2....then the no. of matches would have been nCr...
i.e., selecting r out of n players to play a match....

If each participant of a chess tournament plays exactly 2 game with each of the remaining participants=just multiply nC2 with 2..

You can also approach the question with combinations.

Let's say there are n people playing chess. You need to select 2 people from n people.

Number of ways of doing that is
nC2 = n!/(n-2)!2!

Reducing the above equation you get
n(n-1)/2 number of ways.

Now, n(n-1)/2 = 153
--> n(n-1) = 306

Try different options that fit the criteria. Quick method is to notice that the multiplication of n * n-1 should give a units digit of 6.

Answer D = 18, (18*17 = 306)




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please simplify the above equation in more steps.

Thank you so much. I really appreciate your help sir.

SandhyAvinash

You're very welcome. One last thing.

When you "flag" someone by typing the @ symbol, use it before THEIR name (not yours). All these rules can get confusing. You are very polite, which I appreciate. Cheers!
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Oh thank you i got you genxer123

using your approach,
we can instead save time and see which units digit gives 3 in the end.. 17*9 gives a 3 (dividing by 2 before multiplying the bigger terms) therefore 18 (D) is the answer

My approach was

if there are x players then and each player plays with every other participant, then the each player will play (x-1) games.

Therefore the total number of games = {x(x-1)}/2 as Player A playing with player B is same as Player B playing with Player A.

therefore

x(x-1)*1/2 = 153
x (x-1) = 306, => x=18

P.s :here is it is easy to spot that from the options that with x =18 you will get the last digit as 6. saves on those additional 5 seconds.

Correct Option : D - 18

Method : Combination :
= [n!] / [2x (n-2)] = 153
= [n!] / [(n-2)] = 306

Method 1 :
A - 14+13+12......3+2+1 = 105 x 2 = 210 - Wrong
B - 15+14+13+12......3+2+1 = 120 x 2 = 240 - Wrong
C - 16+15+14+13+12......3+2+1 = 136 x 2 = 272 Wrong
D - 17+16+15+14+13+12......3+2+1 = 105 x 2 = 153 x 2 = 306 - Correct
E - 18+17+16+15+14+13+12......3+2+1 = 171 x 2 = 343 - Wrong

Method 2 :
14x15 - Wrong
15x16 - Wrong
16x17 - Wrong
17x18 - Correct
18x19 - Wrong