GMAT Club World Cup 2022 (DAY 4): How many arch-enemies does Ronaldo

Question

How many arch-enemies does Ronaldo have ?

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

M38-11

Bunuel · Accepted Answer

GMAT CLUB Official Explanation:

How many arch-enemies does Ronaldo have ? 
 
(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
 
Say Ronaldo has  n  arch-enemies:  \{a_1, \ a_2, \ a_3, \ ...a_n\} . Each of  (a_1, \ a_2, \ a_3, \ ...a_n)  has TWO options: either to be included in the list or not. Hence the total number of lists Ronaldo could make is  2^n . 
 
But notice that, this number will include lists with 1 enemy, lists with 2 enemies and an empty list with 0 enemies in it. Since the lists should contain  at least 2 subordinates , then we should subtract all the lists containing only 1 enemy, lists with 2 enemies and an empty list from  2^n .
  
The number of lists with 1 enemy is  n :  \{a_1, \ 0, \ 0, \ 0, \ ..., \ 0\} ,  \{0, \ a_2, \ 0, \ 0, \ ..., \ 0\} ,  \{0, \ 0, \ a_3, \ 0, \ ..., \ 0\} , ...,  \{0, \ 0, \ 0, \ ..., \ a_n\} .
 
The number of lists with 2 enemies is  C^2_n = \frac{n!}{2!(n-2)!}=\frac{(n-1)n}{2} .
 
The number of lists with 0 enemies (empty list) is 1:  \{0, \ 0, \ 0, \ ..., \ 0\} 
  
So, we have that the number of lists with at least 3 arch-enemies is  2^n - n - \frac{(n-1)n}{2}-1 
  
We are told that  469 < 2^n - n - \frac{(n-1)n}{2}-1 < 1069  . 
 
Add 1 to all three parts:  470 < 2^n - n - \frac{(n-1)n}{2} < 1070  .
 
Now, test some numbers. We know that  2^{9}=512  and  2^{10}=1024 , so  n  should be somewhere in that neighborhood: 
 
If  n = 9 , then  2^n - n - \frac{(n-1)n}{2}=512-9-36=467 . Not good!
 
If  n = 10 , then  2^n - n - \frac{(n-1)n}{2}=1024-10-45=969  Good!
 
If  n = 11 , then  2^n - n - \frac{(n-1)n}{2}=2048-11-55=1982 . Not good!
  
Sufficient.
 
(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
 
Apply the same logic as in (1), the difference being that now we are interested in the lists containing  at least 2  enemies instead of the lists containing  at least 3  enemies:
  
 511 < 2^n - n - 1 < 1025  
  
By testing the values, we can get that the above holds true only for  n=10 . 
 
Sufficient.

Answer: D

Kinshook · Answer

Asked: How many arch-enemies does Ronaldo have ?

Let the number of arch-enemies Ronaldo have be n.

(1+1)^n = 2^n = nC0 + nC1 + nC2 + .... + nCn

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).

The number of lists with at least 3 of his arch-enemies = nC3 + nC4 + ... + nCn = 2^n - nC0 - nC1 - nC2 = 2^n - 1 - n - n(n-1)/2
469 < 2^n - 1 - n - n(n-1)/2 < 1069
If n=9; 2^9 - 1- 9 - 36 = 512 - 46 = 466 < 469
If n=10; 2^10 - 1 - 10 - 45 = 1024 - 56 = 968; 469 < 968 < 1069
If n=11; 2^11 - 1 - 11 - 55 = 2048 - 67 = 1981 > 1069
n = 10
Number of arch-enemies Ronaldo have = n = 10
SUFFICIENT

(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
The number of lists with at least 2 of his arch-enemies = nC2 + nC3 + nC4 + ... + nCn = 2^n - nC0 - nC1 - nC2 = 2^n - nC0 - nC1 = 2^n - 1 - n
511 < 2^n - 1 - n < 1025
If n=9; 2^9 - 1- 9 = 512 - 10 = 501 < 511
If n=10; 2^10 - 1 - 10 = 1024 - 11 = 1013; 511 < 1013 < 1025
If n=11; 2^11 - 1 - 11 = 2048 - 12 = 2036 > 1069
n = 10
Number of arch-enemies Ronaldo have = n = 10
SUFFICIENT

IMO D

NextGenNerd · Answer

Correct answer : Choice C

I am guessing this one :
But the thought process being - neither of the statements alone can give the exact combination number. In the sense :

Statement 1 gives : since it is at least 3 arch enemies: the combination will go like : xC3 + xC4 + xC5 ........ and on
Statament 2 would work in a similar fashion: the combination will go like : xC3 + xC4 + xC5 ........ and on

Using both statement 1 and 2 , we can intersect the range of number to a single number and hence figure out the arch enemy count . Hence option C is the correct choice

Iotaa · Answer

1. not sufficient
2. not sufficient.

Combine: S1 tells us about at least 3, which means 3,4,5,6........ & S2 tells us about at least 2, which means 2,3,4,5.......

Subtract at least 3 from at least2, we will get the value for the exact 3. From here we can find rest.

Hence C is the answer

gmatophobia · Answer

Let number of arch-enemies that Ronaldo has = n

Number of ways Ronaldo can select atleast one arch-enemies =  2^n - 1

Number of ways Ronaldo can select atleast exactly one arch-enemy = n

Number of ways Ronaldo can select atleast exactly two arch-enemies = nC2 =  \frac{n(n-1)}{2}

Statement 1

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).

Number of ways Ronaldo can select atleast one arch-enemies = Total number of ways - (none - exactly one - exactly two)
=   2^ n - (1 + n + \frac{n(n-1)}{2})

469 <   2^ n - (1 + n + \frac{n(n-1)}{2})  < 1069

Simplifying

936<   2^ {n+1} - n^{2}-n  < 1069

The only value of n that satisfies this inequality = 10

Therefore A is sufficient

(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

511 <   2^n - (1 + n )  < 1025

511 <   2^n - 1 - n  < 1025

512 <   2^n - n  < 1026

The only value of n that satisfies this inequality = 10

IMO D

false9 · Answer

Feedbacks are welcome.

soudipsengupta · Answer

Pick up reasonable numbers to test cases. Let n be the number of arch enemies.

S1
"Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter)."
Start with n=9, we have 9C3+9C4+9C5+9C6+9C7+9C8+9C9= 466 < 469.. So n cannot be 9
Next with n=10, we have 10C3+10C4+10C5+10C6+10C7+10C8+10C9+10C10 = 968 which lies between 469 and 1069. Therefore n=10 is a possibility
Next with n=11, we have 11C3+11C4+11C5+11C6+... = definitely greater than 1069. So n cannot be 11.
numbers greater than 11 are not possible.
So n=10 .. SUFFICIENT

S2
"Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter)."
Start with n=9, we have 9C2+9C3+9C4+9C5+9C6+9C7+9C8+9C9=36+466=502 < 511. So n cannot be 9
next n=10, we have 10C2+10C3+10C4+10C5+10C6+10C7+10C8+10C9+10C10=45+968=1013 < 1025. So n=10 is a possibility
next n=11, we have 11C2+11C3+11C4+11C5+11C6+... = definitely greater than 1025. So n cannot be 11.
So n=10 .. SUFFICIENT

Ans D

manish1708 · Answer

Answer: D

How many arch-enemies does Ronaldo have ?

Let the number of arch-enemies be n

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
 Lists with at least 3 arch-enemies = 2^n - nCo - nC1 - nC2
=2^n - 1 - n - n(n-1)/2
As per the statement
469 < 2^n - 1- n - n(n-1)/2 < 1069
When n = 9, 2^n - 1 - n - n(n-1)/2 = 512 - 1 - 9 - 36 = 466
when n = 10, 2^n - 1 - n - n(n-1)/2 = 968
469 < 968 < 1069
when n = 11, the value will be more than 1069
Hence, n = 10 is the only possibility.
Number of arch-enemies = 10
 Sufficient .

(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).
 Lists with at least 2 arch-enemies = 2^n - nCo - nC1
= 2^n - 1 - n
As per the statement 
511 < 2^n - 1 - n < 1025
Again, when n = 9, 2^n - 1 - n = 512 - 1 - 9 = 502 < 511
and when n = 10, 2^n - 1 - n = 1024 - 1 - 10 = 1013
511 < 1013 < 1025
when n = 11, the value will be more than 1025
Hence, n = 10 is the only possibility.
Number of arch-enemies = 10
 Sufficient .

Dwiti88 · Answer

How many arch-enemies does Ronaldo have ?

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

Question Stem : Let Ronaldo's arch enemies = n

We know, nC0 + nC1 + nC2 + nC3 + nC4 +.........+ nCn = 2^n

!1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).

Given, 469 < nC3 + nC4 +.........+ nCn < 1069
469 < 2^n - (nC0 + nC1 + nC2 ) < 1069
Only one value of n satsfies ie, n =10

So, Stmt (1) provides an unique answer (Arch enemies = 10)

(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

Given, 511 < nC2 + nC3 + nC4 +.........+ nCn < 1025
511 < 2^n - (nC0 + nC1) < 1025

Only one value of n satsfies ie, n =10

So, Stmt (2) also provides an unique answer (Arch enemies = 10)

Thus, both statements provide unique answer independently

(D) is the CORRECT answer

Hope this helps..

av1901 · Answer

This is a combinations problem which is created to make us work backwards. Instead of giving us total and asking us for combinations, its given us total possible combinations and asking us to calculate the total.
Basically, we are given a range of values nCr (it's nCr and not nPr because it mentions in both statements that order of names in the list does not matter) and the least possible value for r, and we need to calculate the value of n. 
n here will represent the number of arch enemies that Ronaldo has
Let us dive into the statements now:

Note: 
nC2 can be represented as {n(n-1)}/2
nC3 can be represented as {n(n-1)(n-2)}/6
nC4 can be represented as {n(n-1)(n-2)(n-3)}/24
nC5 can be represented as {n(n-1)(n-2)(n-3)(n-4)}/120

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).

Here, we are given:
469 < nCr < 1069 (least value of r=3)

Case 1 (r=3): 
469 < nC3 < 1069
469 < {n(n-1)(n-2)}/6 < 1069
Basically in this scenario, we can have  multiple values of n  which will satisfy the condition (we can factorize the nearest values to 469 and 1069, and see the possible closest 3 consecutive multiples that can via product lead to numbers between this range, for this it came to 16,17,18 and 19 as values of n). 
 But we cannot have multiple number of arch enemies (n), so can we discard it as not sufficient? Maybe, but we need to be sure because sometimes with such a high range, it is possible if we increase the size of the team, it leads to a single answer

Case 2 (r=4): 
469 < nC4 < 1069
469 < {n(n-1)(n-2)(n-3)}/24 < 1069
Here  n is still 12/13/14  but we can see that we have gone down by 1 value from the previous case. Let us try r=5

Case 3 (r=5):
469 < nC5 < 1069
469 < {n(n-1)(n-2)(n-3)(n-4)}/120 < 1069
Here, we finally get  n=12  (A single answer)

So Ronaldo has 12 arch enemies and we can form 12C5 number of lists with 5 of his enemies on each list

SUFFICIENT

(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

Same logic as in Statement 1
With r=2, we get multiple values of n: {From 33 to 45 both inclusive}
DS statements are usually consistent with both statements, so if this statement is supposed to be sufficient, it should also give a single value with r=5
With r=5, we get n=12, a single answer

SUFFICIENT

Answer - D

kungfury42 · Answer

Let Ronaldo have n arch-enemies. Total number of different lists of arch-enemies possible = Lists containing no arch-enemy + Lists containing 1 arch-enemy + Lists containing 2 arch-enemies + ... + Lists containing all arch-enemies = nC0 + nC1 + nC2 + ... + nCn = 2^n Statement 1: Ronaldo could make more than 469 but less than 1069 different unordered lists with at least 3 of his arch-enemies. Total lists containing at least 3 of his arch-enemies = Total lists - = 2^n - nC0 - nC1 - nC2 Try for n = 8, total lists = 2^8 - 8C0 - 8C1 - 8C2 = 256 - 1 - 8 - 28 = 219 (< 469, Incorrect) Try for n = 9, total lists = 2^9 - 9C0 - 9C1 - 9C2 = 512 - 1 - 9 - 36 = 466 (< 469, Incorrect) Try for n = 10, total lists = 2^10 - 10C0 - 10C1 - 10C2 = 1024 - 1 - 10 - 45 = 968 (> 469 and < 1069, Correct) Try for n = 11, total lists = 2^11 - 11C0 - 11C1 - 11C2 = 2048 - 1 - 11 - 55 = 1981 (> 1069, Incorrect) Try for n = 12, total lists = 2^12 - 12C0 - 12C1 - 12C2 = 4096 - 1 - 12 - 66 = 4017 (> 1069, Incorrect) Thus, only n = 10 satisfies. Statement 1 is sufficient. Statement 2: Ronaldo could make more than 511 but less than 1025 different unordered lists with at least 2 of his arch-enemies. Total lists containing at least 2 of his arch-enemies = Total lists - = 2^n - nC0 - nC1 Try for n = 8, total lists = 2^8 - 8C0 - 8C1 = 256 - 1 - 8 = 245 (< 511, Incorrect) Try for n = 9, total lists = 2^9 - 9C0 - 9C1 = 512 - 1 - 9 = 502 (< 511, Incorrect) Try for n = 10, total lists = 2^10 - 10C0 - 10C1 = 1024 - 1 - 10 = 1013 (> 511 and < 1025, Correct) Try for n = 11, total lists = 2^11 - 11C0 - 11C1 = 2048 - 1 - 11 = 2036 (> 1025, Incorrect) Try for n = 12, total lists = 2^12 - 12C0 - 12C1 = 4096 - 1 - 12 = 4083 (> 1025, Incorrect) Thus, only n = 10 satisfies. Statement 2 is sufficient. Hence, both statements are individually sufficient, option D is our answer.

PRamesh2008 · Answer

How many arch-enemies does Ronaldo have ?

(1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).
(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

Question Stem : Let Ronaldo's arch enemies = n

We know, nC0 + nC1 + nC2 + nC3 + nC4 +.........+ nCn = 2^n

!1) Ronaldo could make more than 469 but less than 1069 different lists with at least 3 of his arch-enemies (the order of the names in the list does not matter).

Given, 469 < nC3 + nC4 +.........+ nCn < 1069 
 469 < 2^n - (nC0 + nC1 + nC2 ) < 1069 
 Only one value of n satsfies ie, n =10

So, Stmt (1) provides an unique answer (Arch enemies = 10)

(2) Ronaldo could make more than 511 but less than 1025 different lists with at least 2 of his arch-enemies (the order of the names in the list does not matter).

Given, 511 < nC2 + nC3 + nC4 +.........+ nCn < 1025
 511 < 2^n - (nC0 + nC1) < 1025

Only one value of n satsfies ie, n =10

So, Stmt (2) also provides an unique answer (Arch enemies = 10)

Thus, both statements provide unique answer independently

(D) is the CORRECT answer

Hope this helps..

VelvetThunder · Answer

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GMAT Club World Cup 2022 (DAY 4): How many arch-enemies does Ronaldo

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