5. A counting problem - Lucy & Anton in the photo (GCSE)

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Question. Lucy and Anton are partying with 8 friends. A photo of 6 people in a line is taken. How many different photos are possible if:
(a) Lucy is always in the photo;
(b) Lucy and Anton are always in the photo;
(c) Only one of Lucy or Anton are in the photo.

 

(Higher GCSE Maths 4-9 by Michael White, E8.1 Listing possible outcomes Q14.)

 

Solution. (a) There are a total of 10 people. 6 people are going to be in the photo and Lucy must be one of them.

 

• Then, we choose 5 people out of 9, which gives ${}_9C_5$.
• Together with Lucy, we shuffle 6 people which gives $6!$.
• This gives: $$ \begin{align} {}_9C_5\times 6!=126\times720=90720. \end{align} $$

 

(b) There are a total of 10 people. 6 people are going to be in the photo, and Lucy and Anton must be included.

 

• Then, we choose 4 people out of 8, which gives ${}_8C_4$.
• Together with Lucy and Anton, we shuffle 6 people, which gives $6!$.
• This gives: $$ \begin{align} {}_8C_4\times 6!=70\times720=50400. \end{align} $$

 

(c) We consider:

• From part (a), the number of possibilities where Lucy is included is 90720.
• We can replace Lucy by Anton and find that the number of possibilities where Anton is included is also 90720.
• The number of possibilities where both Lucy and Anton are in the photo is 50400.
• Thus, the number of possibilities where L or A or both are in the photo is: $$ \begin{align} N({\rm L\;or\;A})&=N({\rm L})+N({\rm A})-N({\rm L\;and\;A}) \\ &=90720+90720-50400 \\ &=131040. \end{align} $$
• The number of possibilities where only L or A in included is: $$ \begin{align} N({\rm L\;or\;A})-N({\rm L\;and\;A})=131040-50400=80640. \end{align} $$

 

 

Alternative: We can also consider Lucy only cases and Anton only cases separately.

 

For Lucy only, we must exclude Anton.

• Then, we choose 5 people out of 8, which gives ${}_8C_5$.
• Together with Lucy, we shuffle 6 people, which gives $6!$.
• This gives: $$ \begin{align} N({\rm L\;only})={}_8C_5\times 6!=56\times720=40320. \end{align} $$

For Anton only, the calculation is symmetric, i.e. we must exclude Lucy.

• Then, we choose 5 people out of 8, which gives ${}_8C_5$.
• Together with Anton, we shuffle 6 people, which gives $6!$.
• This gives: $$ \begin{align} N({\rm A\;only})={}_8C_5\times 6!=56\times720=40320. \end{align} $$

 

Finally, $$ \begin{align} N({\rm L\;only})+N({\rm A\;only})=40320+40320=80640. \quad\checkmark \end{align} $$