A man chooses an outfit from 3 different shirts, 2 different

Good problem.

The first day the man can wear anything:
(Shoes)*(Shirts)*(Pants)
(2/2)*(3/3)*(3/3)

The second day the man must wear the same pair of shoes and not wear the shirt and pants from the day before:
(1/2)*(2/3)*(2/3)

The final day, the man must continue to wear the same pair of shoes and wear the one shirt and one pants that were not worn:
(1/2)*(1/3)*(1/3)

So multiplay all together to get the probability of this happening each consecutive day:

(2/2)*(3/3)*(3/3)*(1/2)*(2/3)*(2/3)*(1/2)*(1/3)*(1/3)=
(1/3)^4

ANSWER: C.

h2polo,

Kudos to you as well, u have been ruling the quant section, soon i see u becoming the moderator.
But keep in mind, u r competing with the best....Bunuel.

Oh no... there is no competing with Bunuel. He solves the unsolvable. Hahaha...

I missed out on the activities in this forum lately and it's again a joy to read Bunuel's answers. Kudos

My question is:

day 2: probability of choosing shoe should be 1 as we can not choose one of the two shoes rather the one that was already chosen on day 1
same for day 3
please explain..

san03: On day 2 (and 3), the probability for the shoes should be 1/2
This is because given a shoe was chosen on day 1, the probability of choosing that same shoe the next day is 1/2.

Aah!! i think I got confused between number of ways and probability.
probability of choosing same shoe= 1C1/2C1
Many thanks

Another way to look at the problem :-
!. Since a single pair of shoes has to be chosen and worn for all the three days,
No. of ways of selecting one pair of shoes from the available two = 2C1 = 2.
2. Now the remaining task for us is to choose different shirts and pants for the three days : -
1st day = 3 X 3
2nd day = 2 X 2
3rd day = 1 X 1
Multiplying all the terms
2C1 X 2 X 2 X 3 X 3 X 1 X 1 (Numerator)

now what will come in the denominator?
day Pants Pair of shoes Shirts
1st 3 2 3
2nd 3 2 3
3rd 3 2 3

That is = 3^3 X 2^3 X 3^3 (Denominator)

When the Numerator is divided by the Denominator, you will get the required answer = 1/(3)^4.
Kudos for the answer!

1st Day - \(\frac{3}{3}\) shirts \(*\) \(\frac{2}{2}\) shoes \(*\) \(\frac{3}{3}\) pants

2nd Day - \(\frac{2}{3}\) shirts, \(*\) \(\frac{1}{2}\) shoes \(*\) \(\frac{2}{3}\) pants

3rd Day - \(\frac{1}{3}\) shirts, \(*\) \(\frac{1}{2}\) shoes \(*\) \(\frac{1}{3}\) pants

Answer is \((\frac{1}{3})^4\)

the probability of choosing a unique set of attire on the fisrt day = 1.
from the second day, if I choose to wear the same pair of shoes(s1) on second and third day, then the probaility = [(1/2)*(2/3)*(2/3)]*[(1/2)*(1/3)*(1/3)]= 1/81.

Similarly,I can choose to wear shoes(S2) on all three days. Here as well the probability = 1/81.

Hence, the total probaility = 1/81 + 1/81= 2/81

3 different shirts, 2 different pair of shoes, and 3 different pants..
For first morning, Probability of selecting 1 shirt, 1 pair of shoes, and 1 pair of pants = 3/3 * 2/2 *3/3 =1
For 2nd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 2/3 * 1/2 * 2/3 = 2/9 (2 fresh shirts left and 2 fresh pair of pants left)
For 3rd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 1/3 * 1/2 * 1/3 = 1/18 (1 fresh shirt left and 1 fresh pair of pants left)

So, probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated = 1 * 2/9 * 1/18 = \(\frac {1}{3^4}\)

Answer C