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Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts colored red, pink, blue and white respectively. Arun dislikes the color red and Shweta dislikes the color white. Gulab and Neel likes all the colors. In how many different ways can they choose the shirts so that no one has a shirt with a color he or she dislikes?

  1. 21
  2. 18
  3. 16
  4. 14

I'm totally confused here. Since, Gulab and Neel likes all the colors total number of combination = $4*4 = 16$. Since, Arun and Shweta dislikes 1 color each so there total combination = $3*3 = 9$. So the total ways in which no one choose a shirt no one likes = $16*9 = 144$. Which is way to much high then any of the given options.

3 Answers 3

2

It is easier if you start with the hard assignments. Arun has three choices, but if s/he chooses white the downstream situation is different. So Arun has two non-white choices, then Shweta has two choices, Gulab has two, and Neel is stuck for eight possibilities. Otherwise, Arun chooses white, then Shweta has three, Gulab two, and again Neel is stuck for six more. The total is fourteen.

1

There are $4!$ combinations in total.   Of these $3!$ give Arun a disliked shirt, $3!$ do so for Swheeta, and $2!$ do so for both.

By the Principle of Inclusion and Exclusion we have $4!-2\cdot 3!+2!$, or $14$.

0

We can solve using inclusion-exclusion:

Total possibilities $= 4! = 24$

No.of ways 'arun chooses red' or 'shwetha chooses white'$=$ no.of ways 'arun chooses red'+no.of ways 'shwetha chooses white' -no.of ways 'arun chooses red' and 'shwetha chooses white'$=6+6-2 = 10$

required $=24-10=14$