A playgroup is made up entirely of n pairs of siblings, including the

Question

A playgroup is made up entirely of n pairs of siblings, including the siblings Adam and Josh. 4 members of the playgroup are chosen to represent it in a competition. What is the value of n?

(1) The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
(2) The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5

SaquibHGMATWhiz · Accepted Answer

Solution:      
  Pre Analysis:  
  Total pairs  = n  
 Total people  = 2n  (including Adam and Josh) 
 4 members are chosen from this 2n people  ^{2n}C_{4}  ways 
 We are asked the value of  n

Statement 1:   The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
  Since Adam and Josh are already chosen, we just need to choose other 2 members from  2n-2  remaining members 
 Which can be done in  ^{2n-2}C_{2}  ways 
 Thus according to this statement,  \frac{^{2n-2}C_{2}}{^{2n}C_{4}}=\frac{2}{5} 
 ⇒\frac{\frac{(2n-2)!}{(2n-2-2)!	imes 2!}}{\frac{(2n)!}{(2n-4)!	imes 4!}}=\frac{2}{5} 
 ⇒\frac{\frac{(2n-2)!}{(2n-4)!	imes 2}}{\frac{2n	imes (2n-1)	imes (2n-2)!}{(2n-4)!	imes 24}}=\frac{2}{5} 
 ⇒\frac{12}{2n(2n-1)}=\frac{2}{5} 
 ⇒2(4n^2-2n)=60 
 ⇒8n^2-4n-60=0  
 Since sum of roots is positive and product is negative, we can infer that  we will get one negative and one positive value of n from  8n^2-4n-60=0   
 Ignoring the negative value of n, we will get the required value of n 
 Thus,  statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2:   The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5
  Choosing 2 pairs from n pairs  = ^{n}C_2  
 Thus according to this statement,  \frac{^{n}C_{2}}{^{2n}C_{4}}=\frac{1}{5} 
 ⇒\frac{\frac{(n)!}{(n-2)!	imes 2!}}{\frac{(2n)!}{(2n-4)!	imes 4!}}=\frac{1}{5} 
 ⇒\frac{\frac{n	imes (n-1)	imes (n-2)!}{(n-2)!	imes 2}}{\frac{2n	imes (2n-1)	imes (2n-2)	imes (2n-3)	imes (2n-4)!}{(2n-4)!	imes 24}}=\frac{1}{5} 
 ⇒\frac{n(n-1)}{2n	imes (2n-1)	imes (2n-2)	imes (2n-3)}=\frac{1}{60} 
 ⇒\frac{1}{4(2n-1)(2n-3)}=\frac{1}{60} 
 ⇒4n^2-8n-12=0  
 Since sum of roots is positive and product is negative, we can infer that  we will get one negative and one positive value of n from  4n^2-4n-12=0   
 Ignoring the negative value of n, we will get the required value of n 
 Thus,  statement 2 alone is also sufficient    
 
Hence the right answer is    Option D

ASRV · Answer

Can someone share the solution for this, please.

gullyboy09 · Answer

We don't really need to solve the equation. Given it's one variable equation, it will always give us n value. Now even if equation comes cubic, or even higher power, it can still be solved by computers, so we have enough information to know n. Question is not asking what is n but asking can we find n using the information given. It would have been relevant if we had two variables and for some reason we are not able to remove them from the fraction or convert into one variable.

On checking if real roots exist or not, we don't need to check that either. Given statement 1 is "given", meaning that situation does exist, then we will have solution definitely, equation cannot have all imaginary roots.

On checking if real roots exist or not, we don't need to check that either. Given statement 1 is "given", meaning that situation does exist, then we will have solution definitely, equation cannot have all imaginary roots.

A playgroup is made up entirely of n pairs of siblings, including the

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