1st team could be any of 2 guys... there would be 4 teams (a team of A&B; is same as a team of B&A;)... possible ways 8C2 / 4. 2nd team could be any of remaining 6 guys. There would be 3 teams (a team of A&B; is same as a team of B&A;)... possible ways 6C2 / 3 3rd team could be any of remaining 2 guys... there would be 2 teams (a team of A&B; is same as a team of B&A;). Possible ways 4C2 / 2 4th team could be any of remaining 2 guys... there would be 1 such teams... possible ways 2C2 / 1 total number of ways... 8C2*6C2*4C2*2C2 ------------------- 4 * 3 * 2 * 1 = 8*7*6*5*4*3*2*1 -------------------- 4*3*2*1*2*2*2*2 = 105 (ANSWER)...
Got it....
The mistake that I did was taking the order of the teams into account, which is not required for this question.
What this means is, Say we have 8 people numbered from 1-8. One set of the four teams: (1,2), (3,4), (5,6), (7,8 ) Now even if we change the order of the teams to (5,6), (3,4), (1,2), (7,8 ), it does not change the number of TEAMS, that, remains exactly same, that is why we need to divide the whole answer by 4!, taking into account the 4 Teams.
sausi, I am sorry but this might seem silly. But i am a little lost with the explanation. Can you detail it? Thanks in advance.
This problem is essentially asking us to find the number of ways to choose any 5 students out of the available 20, and, since there is only one way in which these 5 students will be arranged in the queue, we dont need to do anything else.
So, simply, 10C5 should give us the answer = 15504
There are 20 students in a class, such that no two students are of same height. In how many ways can any 5 students stand in queue in increasing order of height. The reason for doing so was Once we have selected 5 students from the class, and we know that no 2 students have the same height, if we have to make these 5 students stand in increasing order of height, there will only be ONE way of doing so. Therefore, the solution will be= No of ways of selecting 5 students from a class of 20 * 1(No of ways of making them stand in increasing order of height) = 20C5*1 = 15504 (There was a mistake in the earlier explanation, it should be 20C5, not 10C5)
There are 20 students in a class, such that no two students are of same height. In how many ways can any 5 students stand in queue in increasing order of height. The reason for doing so was Once we have selected 5 students from the class, and we know that no 2 students have the same height, if we have to make these 5 students stand in increasing order of height, there will only be ONE way of doing so. Therefore, the solution will be= No of ways of selecting 5 students from a class of 10 * 1(No of ways of making them stand in increasing order of height) = 20C5*1 = 15504 (There was a mistake in the earlier explanation, it should be 20C5, not 10C5)
The infinite sequence a1, a2,, an, is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80
The infinite sequence a1, a2,, an, is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80
Consider a string of n 7s, 7777....77, into which + signs are inserted to produce an arithmetic expression. For example, 7 + 777 + 7 = 791 could be obtained from five 7s in this way. For how many values of n is it possible to insert + signs so that the resulting expression has value 7000?
Consider a string of n 7s, 7777....77, into which + signs are inserted to produce an arithmetic expression. For example, 7 + 777 + 7 = 791 could be obtained from five 7s in this way. For how many values of n is it possible to insert + signs so that the resulting expression has value 7000?
(1) 105 (2) 106 (3) 108 (4) 109 (5) 111
We have to work the problem backwards starting from the big number (777) to small number (7)
111a+11b+c = 1000 3a+2b+c = a unique number say 'N'
Now 9b+ 108a = 1000-N
N = 9x+1 N = 10, 19, 28 ... 1000 As 28
I doubt whether this is a question which we should really bother too much about, the checking of values of N by putting them in the eqn does not make any sense. I will never check more than 3 values. Dont know why this question has such a solution.
Can you pls explain how the following has been arrived @
1) 3a+2b+c = a unique number say 'N'
2)N = 9x+1
-Deepak.
a, b,c are all used ofr counting the no of 7's 3a (counts the no of 7's used by 777) 2a (counts the no of 7's used by 77) a (counts the no of 7's used by 7)
So 777a+77b+7c=7000 (@Suez, please try to detail the steps for others to understand) Thus, 111a+11b+c=1000
Now we need to find N, which will be = 3a+2b+c Thus, substracting the 2 eqns, we get, 108a+9b=1000-N or, 9(12a+b)=1000-N
now taking 12a+b = x 9x = 1000-N, where N varies from 1 to 1000
This is where i will stop, i dont think i will check all the values which satisfy this eqn, @Suez: how can anyone be sure that after 37, there is no value which does not satisfy this eqn?
yes, indeed it would be 74