A,B,C,D, E, F, G can stand in 7! ways .requirement is : CDB - can be arranged in 3! ways but only one is acceptableHence total ways = 7!/3!PS: galat hone ke pure assaar hain
Dekhte hain baki janta kya bolti hai tumhare asaar ke bare me:)
The equation |z-1| - |z-2| + |z-4| = k has exactly r real solutions for some real k. Then which among the following relations between k and r can not be true?
(a) k/r = 3/5 (b) k = r (c) k/r = 3/2 (d) k/r = 5/3 (e) k = r-1
Good job Amresh and Estallar12.Try this one now...The equation |z-1| - |z-2| + |z-4| = k has exactly n real solutions for some real r. Then which among the following relations between k and r can not be true?(a) k/r = 3/5 (b) k = r (c) k/r = 3/2 (d) k/r = 5/3 (e) k = r-1Team BV - Vineet
Good job Amresh and Estallar12.Try this one now...The equation |z-1| - |z-2| + |z-4| = k has exactly n real solutions for some real r. Then which among the following relations between k and r can not be true?(a) k/r = 3/5 (b) k = r (c) k/r = 3/2 (d) k/r = 5/3 (e) k = r-1Team BV - Vineet
k/r = 3/2 nahin aa raha hai... Edit: n aur r mein confusion ho raha hai....please clarify....maine k aur n mein nikala actually....r ka importance kuch samajh nahin aaya
Good job Amresh and Estallar12.Try this one now...The equation |z-1| - |z-2| + |z-4| = k has exactly n real solutions for some real r. Then which among the following relations between k and r can not be true?(a) k/r = 3/5 (b) k = r (c) k/r = 3/2 (d) k/r = 5/3 (e) k = r-1Team BV - Vineet
k/r = 3/2 nahin aa raha hai...Edit: n aur r mein confusion ho raha hai....please clarify....maine k aur n mein nikala actually....r ka importance kuch samajh nahin aayaregardsscrabbler
In how many ways can A,B,C,D, E, F, and G stand in queue such that D is always behind B and C is always behind D?Team BV - Vineet
Correct Answers: - 840 ways
1st method: - Out of 7 persons, we need to choose 3 and we can arrange them only 1 way. (like C>D >B) and remaining 4 persons, we can arrange 4! ways. Total number of ways = C (7, 3) x 1 x 4! = 840.
2nd method: - CDB can be arranged in 3! ways but only 1 combination is in our favour. So, Total Combinations (without any condition) = 7! Required combinations = 7!/3! = 840.
k/r = 3/2 nahin aa raha hai...Edit: n aur r mein confusion ho raha hai....please clarify....maine k aur n mein nikala actually....r ka importance kuch samajh nahin aayaregardsscrabbler
Yes scrabbler. There was a typo, which i have edited. Check the problem now.
In a multinational company, the employee ID is a three-digit number (first digit is non-zero). However, no two employees can have IDs that are identical when written in reverse order (123 and 321 are identical and hence only one of them can exist). What is the maximum number of employees that can use this coding system?
