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
