In a clothing store, there are six different colored neckties (red, or

Hi,

6 ties and 6 shirts...
red tie can take any of 6 shirts..
orange can take any of the remaining 5 shirts
yellow any of remaining 4..
and so on till last indigo chooses the 1 remaining..

Total ways= 6*5*4*3*2*1= 720

out of this 720, ONLY 1 way will have same colour tie and shirt..

prob = 1/720
E
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Slightly different way to think about it...

First box can take from 6 shirts and 6 ties (6*6 possibilities)
Second box can take from 5 shirts and 5 ties (since one of each was already used in the first box) (5*5 possibilities)
Third box can take from 4 shirts and 4 ties (4*4 possibilities)
etc
Total ways to pack those boxes: (6!)^2
But this counts different arrangement of boxes (1 blue shirt and 1 red tie in box 1 is different from 1 blue shirt and 1 red tie in box 2)... So we need to un-arrange, i.e. divide by 6!.

Total ways to pack the boxes = 6! = 720
Only one way to have RR, OO, YY, GG, BB, II

Answer: E

Also, if we had to guess, we could probably pick E
A. 719/720 ---> way too big
B. 1/120 ---> maybe
C. 2/233 ---> 233 in the denominator doesnt look familiar (not a square of an integer and not a factorial). also, we're looking for a '1' in the numerator
D. 3/543 ---> same logic as for C. I dont know how they could have 543 in denominator. also, we're looking for a '1' in the numerator
E. 1/720 ---> probably. and because of the complementary option A, i would think A is the trap and E is the correct answer

In a clothing store, there are six different colored neckties (red, orange, yellow, green, blue, and indigo) and six different colored shirts (red, orange, yellow, green, blue, and indigo) that must be packed into boxes for gifts. If each box can only fit one necktie and one shirt, what is the probability that all of the boxes will contain a necktie and a shirt of the same color?

A. 719/720
B. 1/120
C. 2/233
D. 3/543
E. 1/720

Total probability is 6! = 720

Now the answer choices are interesting here. Only the options A and E have denominator 720. The option B has 120 so we can hold on to this option for now. Option A has probability of 719/720 which is high and less probably answer choice. We can eliminate options C and D based the denominator. We are left with option B and E.

Of the remaining options , only one box will have shirt and tie in same color.

Ans: E
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For the red necktie, there is a 1/6 chance it will be packed with the red shirt. Given that the reds have been packed together, for the orange necktie, there is a 1/5 chance it will be packed with the orange shirt. Given that the oranges have been packed together, for the yellow necktie, there is a 1/4 chance it will be packed with the yellow shirt. From this point, we see that there is a 1/3 chance the green necktie and shirt will be packed together, a 1/2 chance the blue necktie and shirt will be packed together and, finally, a 1/1 chance the indigo necktie and shirt will be packed together. Therefore, the overall probability is 1/6 x 1/5 x 1/4 x 1/3 x 1/2 x 1/1 = 1/6! = 1/720.

Answer: E
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Once the six shirts have been distributed among 6 boxes, one shirt per box:

P(the first box receives its matching tie) = 1/6 (Of the 6 ties, only 1 is a match.)
P(the next box receives its matching tie) = 1/5 (Of the 5 remaining ties, only 1 is a match.)
P(the next box receives its matching tie) = 1/4 (Of the 4 remaining ties, only 1 is a match.)
P(the next box receives its matching tie) = 1/3 (Of the 3 remaining ties, only 1 is a match.)
P(the next box receives its matching tie) = 1/2 (Of the 2 remaining ties, only 1 is a match.)
P(the next box receives its matching tie) = 1/1 (The 1 remaining tie must be a match.)
To combine these probabilities, we multiply:
1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/720


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Conceptual approach:

https://ibb.co/MBRBNL5
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Probability = Favorable/Total

Total = 6C1 * 6C1
Favorable = 6C1 * 1C1

P = 6C1*1C1/6C1*6C1 = 6!/6!*6! = 1/6! = 1/720 (E)

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