During a season in a certain baseball league every team plays every other team in the league 10 times. If there are 10 teams in the league, how many games are played in the league in one season?
a 45 b 90 c 450 d 900 e 1000
Well, here's my reply. Correct answer is C.
If there are N teams playing each other once, total number of games played = NC2 (two teams selected out of N to play a game). If 10 such games are played for every combination, total games = 10 * NC2.
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In the question given, N = 10. Therefore, total number of games = 10 * 10C2 = 10 * 45 = 450.
P.S: I guess one of the guys got it correct finally. But it is important to not be in a hurry or get confused. Attention must be paid to concepts. It helps us improve speed and accuracy.
This one is for the more serious minded guys....a history of evolution of Geometry. Dwells on:
What is Geometry? What are the different types of Geometry? Why did man think of these different types of Geometry? What were the problems that gave rise to these different studies? What practical applications did the studies lead to?
Q. 2/3 rd's of the roads from A to B are at least 5 miles long and 1/4 th of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
Q. 2/3 rd's of the roads from A to B are at least 5 miles long and 1/4 th of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
(a) 1/6 (b) 1/4 (c) 2/3 (d) 3/4 (e) 11/12
Thanks in advance.
Is ans A? They are two independent event so probability for them to happen together is multiplication of individual event P(A1,A2)=P(A1)*P(A2)
Q. 2/3 rd's of the roads from A to B are at least 5 miles long and 1/4 th of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
(a) 1/6 (b) 1/4 (c) 2/3 (d) 3/4 (e) 11/12
Thanks in advance.
Answer is 3/4. Option D.
(2/3)rds of the roads from A to B are at least 5 miles long => >= 5 Miles.
(1/3)rds of the roads from B to C are at least 5 miles long => >= 5 Miles.
If we randomly pick a road from A to B AND randomly pick a road from B to C, the probability that at least one of the roads is >= 5 miles is when
1. Road from A to B >=5 Miles but not that from B to C + 2. Road from A to B = 5 Miles +
(when either of the roads is at least 5 Miles long)
3. Road from A to B >=5 Miles and also that from B to C >= 5 Miles (when both are at least 5 Miles long)
The function F is defined for all positive integers n by the following rule: f(n) is the number of positive integer each of which is less than n, and has no positive factor in common with n other than 1. If p is any prime number then f(p)=
The function F is defined for all positive integers n by the following rule: f(n) is the number of positive integer each of which is less than n, and has no positive factor in common with n other than 1. If p is any prime number then f(p)=
a)p-1 b)p-2 c)(p+1)/2 d)(p-1)/2 e) 2
even i would go with answer A: p-1 what is the oa?
The function F is defined for all positive integers n by the following rule: f(n) is the number of positive integer each of which is less than n, and has no positive factor in common with n other than 1. If p is any prime number then f(p)=
Q. How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?
(a) 2 (b) 3 (c) 4 (d) 6 (e) 12
Lets see the total arrangements possible-
Let the students sit on 1st and 3rd desks so the total number of ways in which 2 students can sit is 2 ways i.e. the 1st student either sits on 1st desk or on 3rd desk.
Now let the students sit on 2nd and 4th desks so the total number of ways in which 2 students can sit is 2 ways i.e. the 1st student either sits on 2nd desk or on 4th desk.
therefore, total numbers of ways 2 students can sit across 4 desks one empty desk between them = 2+2= 4
Now we have to also consider a case when there are 2 desks empty between the two students and i.e. only possible when one of the students occupy either the 1st desk or 4th desk, hence total number of ways to occupy 1st and 4th desk between 2 students= 2
Hence, total number of ways with at least one empty desk between the two students = 4+2 =6
Q. How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?
(a) 2 (b) 3 (c) 4 (d) 6 (e) 12
Let the seats be 1,2,3 and 4 and the students be A and B. If A seats at 1, B can seat at 3 or 4, so two ways. If A seats at 2, B can seat at 4 only, So one way. If A seats at 3, B can seat at 1 only, so one way. If A seats at 4, B can seat at 1 or 2, so two ways.
Q. How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?
(a) 2 (b) 3 (c) 4 (d) 6 (e) 12
Let there be seats 1,2,3,4 and x and y be two persons. Now, if x sits on seat 1 then y can sit on 4th or vice-versa then there are 2 ways. if x sits on 2nd seat and y on 4th seat or vice versa then there are 2 ways if x sits on seat 3rd then y can only sit on seat 1 or vice versa then there are 2 ways. now adding up all 2+2+2, we get 6 ways.
Q. How many different ways can 2 students be seated in a row of 4 desks so that there is always at least one empty desk between the students?
(a) 2 (b) 3 (c) 4 (d) 6 (e) 12
Yes, the answer is indeed 6. Option D.
Here is another way of doing it:
The total number of ways of seating 2 people in a row of 4 desks = 4P2 (Permutations or Arrangements) = 12.
Now, to find out the number of ways in which they can be seated so that there is atleast one seat between them = 12 - (No. of ways in which they are always seated together).
No. of ways they are always seated together = 6 (ABXX, BAXX, XABX, XBAX, XXAB, XXBA).
My take was since the two set of arrangement above are identical (you just flip flop A and B) and the questions doesn't say two distinct students, I divided 6 by 2.
My take was since the two set of arrangement above are identical (you just flip flop A and B) and the questions doesn't say two distinct students, I divided 6 by 2.
@m_varun,
The question was about the number of ways of seating people (which is an arrangement; in other words, permutations) not about selection of people (which is a combination). Therefore, for each particular selection and arrangement, there is indeed another arrangement or way of seating people (reversing the positions). Total = 6.
