Let Q be the set of permutations of the sequence 1,2,3,4,5 for which t

Question

Let Q be the set of permutations of the sequence 1,2,3,4,5 for which the first term is not 1. A permutation is chosen randomly from Q. The probability that the second term is 3, in lowest terms, is a/b. What is a+b?

A) 5
B) 6
C) 11
D) 16
E) 19

A) 5
B) 6
C) 11
D) 16
E) 19

mikemcgarry · Answer

Dear manpreetsingh86 , Wow! This is a good juicy hard question! I like it! :-)

For a discussion of probability and counting techniques, see: https://magoosh.com/gmat/2013/gmat-prob ... echniques/ OK, so first, we figure out the size of Q. As always, start counting with the most restrictive case. Since 1 cannot be in the first term, there are four possibilities for the first term; then 4 left for the second term; then 3, then 2, then 1. N = 4*4*3*2*1 = 16*6 = 96 That's the denominator of our big probability fraction, the total number of possible permutations in Q Now, for the numerator, the second term is the most restrictive: we have to put 3 in the second term. That's only 1 option in that slot. That leaves only 3 possible choices for the first term, which can be neither 3 nor 1. Then, once the first two are selected, we have 3 choices for the third term, 2 for the second, and 1 for the fifth. N = 1*3*3*2*1 = 18 Probability = 18/96 = 3/16 3 + 16 = 19. Choice (E) . That was a fun problem! If anyone has any questions, please let me know. Mike :-)

KarishmaB · Answer

Use a bit of logic of symmetry to solve this question without any calculations.

Q would include all such sets {2, 3, 1, 4, 5}, {3, 1, 2, 4, 5}, (4, 3, 2, 5, 1), {5, 3, 2, 1, 4} etc - starting with 2, with 3, with 4 or with 5.
Note that 1/4th of them will start with 3 so we are left with the rest of the 3/4th sets.

Now, in these 3/4th sets which start with either 2 or 4 or 5, 3 could occupy any one of the 4 positions - second, third, fourth or fifth with equal probability. So we need 1/4th of these sets.

Probability that 3 is the second element of Q = (3/4)*(1/4) = 3/16

So a+b = 3+16 = 19

Answer (E)

chetan2u · Answer

hi..
the total possible sequence =5!, out of which there are 4! where 1 is the first term..
so set of permutations = 5!-4!=96..

now putting the restrictions as imposed..
2nd position by 3.. so 1 way
1st by 2,4,5 so 3 ways
3rd by 1,2,4,5 minus the one occupying 1st position. so 3 ways
4th by remaining two(1,2,4,5 minus (the one occupying 1st position and the one occupying 2nd posn) so 2 ways
5th by the sole remaining digit...
total ways..3*1*3*2*1=18..
so prob=18/96=3/16 in lowest terms. therefore 3+16 =19..ans E

manpreetsingh86 · Answer

Here is an another solution, may be slightly complex.

now, since 1 can't be the first term, therefore 1 can occur at any of the remaining 4 places. So, the probability that 1 can occur at the second position is 1/4.

Let's assume that x be the probability of 3 to occur at the second position, now this probability will be same for all the other numbers except 1. Also, one number is definitely going to occur at position 2, therefore, we have

1/4 + 4x = 1

or 1 + 16x =4

or x = 3/16

so the required answer is 3 + 16 =19.

chetan2u · Answer

hi manpreet,
 a different approach altogether.. i am sure if someone gets hang of it, a good amount of time can be saved..

mvictor · Answer

I used to hate such types of questions...yet now I cracked this bad boy in under 2 minutes...
so...
to find the denominator -> can't use 1 as the first digit
we thus have 4x4x3x2x1 options, or 96.

we need 3 to be the second digit...thus, we would have only 1 option for second digit, and 3 options (can't use 1 and 3) for first digit. for the third digit we can use 3 options (because from 5, we used 3 and some other digit), then fourth 2 options, and fifth one option.

3x1x3x2x1

now, we are told that 3x1x3x2x1/4x4x3x2x1 is a/b
let's do some cancelling
3/4x4 = 3/16 where 3 is a, and 16 is b
we are asked a+b =3+16=19

mvictor · Answer

wow, that's an awesome approach!!!

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Let Q be the set of permutations of the sequence 1,2,3,4,5 for which t

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