winnerhere wrote:A class has 100 students with roll numbers from 1 to 100.Six students are selected one after another at random to form a teamto take part in a quiz competition.Find the probability that the roll numbers of the first three students are in the ascendig order and the roll numbers of the last four students are in the descending order.
1. 1/720
2. 1/360
3. 1/180
4. 1/72
5. 1/36
Hi! Are you sure that the question is posted correctly? I'm getting a different answer to the ones listed.
The first thing I thought when I read the question was "geez, that's super complicated and will require a lot of calculations!" Fortunately, my second thought (right on the heels of the first) was "wait - this is a GMAT question, it CAN'T require too much math!"
We can answer this question using basic permutations. Let's look at the two separate parts.
First, we want the first 3 numbers to be in ascending order. For 3 objects, there are 3! (3*2*1 = 6) different ways in which they can be arranged. If you were to write them out:
123
132
213
231
312
321
Only 1 of those 6 is in ascending order, therefore the probability that the first 3 are in ascending order is 1/6.
Second, we want the last 4 (I think this is the typo in the question) numbers to be in descending order. Applying the same approach, there are 4! (4*3*2 = 24) different possible arrangements, only 1 of which will go 6543. Therefore, the probability that the last 4 are in descending order is 1/24.
Since we want the probability of MULTIPLE events occuring, we MULTIPLY the individual results:
(1/6) * (1/24) = 1/144
which, sadly, isn't among the choices.
Now, it's certainly possible that I've made a mistake. However, my hubris shining through, I'm guessing that the end of the question should have read:
"and the roll numbers of the last
three students are in descending order."
In which case the answer would be:
(1/6) * (1/6) = 1/36... choose (5).
