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Permutation and combination

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by parveen110 » Tue Feb 04, 2014 5:56 am
A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dice. How many ways are possible of having one before six?
a.120
b.360.
c.240
d.380
e.280

OA:360
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by Brent@GMATPrepNow » Tue Feb 04, 2014 8:27 am
parveen110 wrote:A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dice. How many ways are possible of having one before six?
a.120
b.360.
c.240
d.380
e.280

OA:360
This question is very similar to this one: https://www.beatthegmat.com/mobster-comb ... 66632.html

Presumably (judging from the OA), a die is rolled 6 times, and in those 6 rolls, we get exactly one 1, one 2, one 3, one 4, one 5, and one 6.
So, some possible scenarios are:
2-4-5-1-3-6
1-2-4-6-3-5
6-1-3-5-2-4
etc

We want to determine the number of arrangements such that the 1 appears before the 6.

If we IGNORE the restriction that the 1 must appear before the 6, we can see that we can arrange the six digits (1,2,3,4,5 and 6) in 6! ways. (using a rule that says we can arrange n unique objects in n! ways)
6! = 720
So we can arrange the 6 digits in 720 ways

IMPORTANT: Notice that, in HALF of those 720 arrangements, the 1 appears before the 6, and in the other HALF of those 720 arrangements, the 6 appears before the 1.
So, there are 360 arrangements in which the 1 appears before the 6
Answer: B

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Tue Feb 04, 2014 8:28 am, edited 1 time in total.
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by GMATGuruNY » Tue Feb 04, 2014 8:27 am
parveen110 wrote:A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dice. How many ways are possible of having one before six?
a.120
b.360.
c.240
d.380
e.280

OA:360
The number of ways to arrange the six digits 1, 2, 3, 4, 5, and 6 = 6! = 720.
In any given arrangement, the probability that 1 comes before 6 is the same as the probability that 6 comes before 1.
Thus:
In 1/2 of the arrangements, 1 will come before 6.
In the other 1/2 of the arrangements, 6 will come before 1.
Result:
The number of arrangements in which 1 comes before 6 = (1/2)(720) = 360.

The correct answer is B.
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by [email protected] » Tue Feb 04, 2014 2:17 pm
Hi parveen110,

Both Brent and Mitch presented the Number Property shortcut that "half" the possibilities are negated because the 1 has to come before the 6 (and not after the 6). If you didn't realize that shortcut, there's still another way to solve the problem; it just takes a little more work.

We're going to track each of the possibilities by using permutation rules. The numbers 2, 3, 4 and 5 can go anywhere, but the placement of the 1 affects where we can put the 6:

_ _ _ _ _ _

Here we have 6 spots (for the six dice rolls).

If the 1 is in the first spot, then the 6 (as well as the other 4 numbers) come after. We have:

1x5x4x3x2x1 = 120 possibilities in which the 1 comes before the 6

If the 1 is in the second spot, then the 6 CAN'T go in the first spot. We have:

4x1x4x3x2x1 = 96 possibilites

If the 1 is in the third spot, then the 6 CAN'T go in the first or second spot. We have:

4x3x1x3x2x1 = 72 possibilities

If the 1 is in the fourth spot, then the 6 CAN'T go in the first, second or third spots. We have:

4x3x2x1x2x1 = 48 possibilities

If the 1 is in the fifth spot, then the 6 MUST be in the last spot. We have:

4x3x2x1x1x1 = 24 possibilities

120 + 96 + 72 + 48 + 24 = 360 possibilities

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by parveen110 » Wed Feb 12, 2014 11:39 pm
Rich, By the same logic one more approach occured to me.

Like you said, the constraint is placed on the outcomes of 1 and 6 only, i.e. 1 should come before 6.
And the other outcomes viz. 2,3,4,5 can occur at any place during six consecutive throws.

so, # of options for 2 to occur at any place is: 6
# of options for 3 to occur at any place is: 5
# of options for 4 to occur at any place is: 4
# of options for 5 to occur at any place is: 3
# of options for 6 to occur at any place is: 1(right most position)
# of options for 1 to occur at any place is: 1(left most position)

combining all the options, 6*5*4*3*1*1=360
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by [email protected] » Wed Feb 12, 2014 11:49 pm
HI parveen110,

Your logic absolutely works because the other 4 numbers (the 2, 3, 4 and 5) can appear anywhere in the line (and then it's just a matter of placing the 1 and 6 according to the "rules").

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