Top_Gun

Well, we need to be lucky, but u don't. Well, u seem like a math wizard. That's why they have left enough leeway in the scoring system to ensure perfect score for u guys. We should rather be worried, we'ld fail to answer some maths cause we can't and also some cause the question might not be a well articulated one.

Anyway, where r u from?

Hmmm... As stickler agreed with u too, i don't have any doubts. When someone with ur intellect and a genious like stickler agrees on a thing, then it must be it.

I agree with MWM.

What's the rear of shah, is it riar???

Congrates. U r right on both of the problems.

don't need for myself to look it up. I believe u guys. I'm feeling myself lucky already. Thanx.

Just consider first the col A. Among the numbers from 1 to 6, how many of them r closer to 4 than to 2. 4,5 and 6, right? so, the probability that a roll is die and out of 6 sides we came up with a 4 or 5 or 6 is 3/6 = 1/2. consider column B. The number that satisfies the condition is 5 and 6. Even, 4 is in equal distance from both 3 and 5. so, the probability is 2/6 = 1/3 So, column A is greater. the numbers that satisf

Well, in that case, i expect some clear cut indications in the question. As no such indications were given in the question, i had to solve the question that way. Ur answer may be right, but it all depends upon what the question makers thought while making the question, not what is meant by the question.

If we are to find all the permutations where S has to come together, we can simply assume all the S's( 4 S's ) to be just one character. Then all S would come together. So, now there r 7 characters( M, P, 4 I's and a single S ). So, number of permutations = 7!/4! Again, the question asks us to find out the numbers where none of the I's would come together. So, we've to just find the number of such combinations where I's come together. Now, likewise, we'ld think that all the I's are also a single I. So, the number of characters is: 4( M, P, only a solitary I and a solitary S ) and all of them r distinct characters. So, number of such characters : 4! So, answer is = 7!/4! - 4!

U r always welcome.

suppose, that person has failed in 2 subjects. Now, given that, there can be how many possible ways to do that? wydiwyg, u said 2. But, wouldn't it be 5c2??? I think, that's a problem in ur solution. In ur way to solve it: Failing in 5 subjs: 1 way. Failing in 4 subjs: 5c4 = 5 ways. Failing in 3 subjs: 5c3 = 10 ways. Failing in 2 subjs: 5c2 = 10 ways. Failing in 1 subjs: 5c1 = 5 ways. So, total number of ways: 1 + 5 + 10 +10 + 5 = 31 ways. So, it's the same result: 31 = 2 ^ 5 - 1 Just that, 2 ^ 5 - 1 was more elegent and less time consuming.

Well, i hardly know any formula about these permutation combination problems. I solve them intuitively. For the letter problem, Just think of each letter. A letter can be one of each 5 letters. So, it can be placed in 5 ways. The second letter too can be placed in one of the 5 boxes. So, two letters can be placed in 5^2 ways. .................. ................... So, 10 letters in 5^10 ways. and..the other one: a person can either fail in a subject or pass. So, he can have 2^5 ways to pass or fail. But, in one of these combinations he passes all the subjects. So, the answer is: 2^5 - 1