Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?(a) 8 (b) 10 (c) 15 (d) 22
Possible ways =2^5=32
Unfavourable= 2 B together, 3 B together, 4 B TOGETHER 2 B together BBR _ _ RBBR_ _RBBR _ _ RBB TOTAL=4+2+2+4=12; Subtract 1 for double counting of BBRBB so total=11 3 B together BBBR_ RBBBR _RBBB TOTAL 5 4 B TOGETHER=1
I think answer should be 22For 0 blue balls---> 1 waysFor 1 blue ball-->5 waysFor 2 blue balls--->15 ways (5 ways for one and 3 ways for the other)For 3 blue balls---> 1 wayFor 4--->not possible
Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?(a) 8 (b) 10 (c) 15 (d) 22
R R R R R - 1 R R R R B - 5 R R R B B - 6 R R B B B - 1
sir main jaise kar raha tha....... 3 log hain to unko 1540 dediye har kisi ko.....ab 1540 bacha to 1540 3 se divide nahi hota........to matlab remainder aata hai...par agar do ke case 2310 ho gaya par ye further 2 se divide ho jata hai to reminder zero aa gaya......isme kuch galat kar raha hoon kya sir????????
Yes, this method works too, but the simpler method is to look at the factors with even powers . As I said, if n divides 4620, n^2 should not divide 4620 -> n^2 should not be in the original list of factors of 4620
In this case, 1 and 2 are in the factors as well as their squares 1 and 4 are. This method is faster, but your method is fine too 😃
@joyjitpal Can you explain the bold part ? BBRRR has total arrangements = 10 ways ! ( 5!/3!2!) So, if we don't want B in adjacent boxes => Total - (when 2 Bs are together) = 10 - (4!/3!) = 6 ?Also, another way _B_B_ , a + b + c = 3 -> a + b' + c = 2 => 4C2 = 6 ways ?How did you get 15 ? R B R B RR R B R BB R B R RB R R B RR B R R BB R R R BWhat else ? I guess, I didn't get the question.PS: I guess, today is my NONE OF THESE options Day
Sorry..apologies
I considered just the 1st case where the ball is in the 1st position and applied the rule to all the balls..guess from your explanation my answer too would be none of these
Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?(a) 8 (b) 10 (c) 15 (d) 22
_ = 1 way _B_ = 5 ways _B_B_ => a + b' + c = 2 => 4C2 = 6 ways B_B_B = 1 way
sir even kyun any even odd power will work if there is an odd power >1....for ex:- divide 27 for d same question....... not possible......3^3........if each gets 9 we can further divide it......in d last case had dere been a factor of d form n^2x+1.....den we wud hav considered it as well..........plz correct me if i m wrong
Yes, so taking your example i.e. 27
n = 3 is rejected right ? Why ? Because 3^2 is a factor of 27 too , that's why
In general 27/n is an integer = k , given in the question, but (k/n) should not be as per the requirement
=> 27/(n^2) is NOT an integer ..
I didn't say take the final power of the prime factors, I said, if there is a possibility that an even power of the factor (n) exists, then reject that n.
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?
(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?(a) 8 (b) 10 (c) 15 (d) 22
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
Select the box with the red and white written on it
Since the box is incorrectly labelled, we can choose one ball from it
If the ball is white then that box contains only white balls and if it is red then only red
From this we can also identify the other 2 boxes since they are also incorrectly labelled
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
c) Red and White ?
If we get Red when we pick from it, => The box marked Red is White, and the other one is Mixed
If we get White, => The box marked White is Red, and the other is mixed.
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
c) Red and White ?If we get Red when we pick from it, => The box marked Red is White, and the other one is MixedIf we get White, => The box marked White is Red, and the other is mixed.
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?(a) White (b) Red (c) Red and White (d) Not possible to determine from a sample of one ball
Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?(a) 8 (b) 10 (c) 15 (d) 22
13 ways No B---1 way 1 B-----5 ways 2 B-----6 ways 3 B-----1 way Total=13 ways
.. Only smileys not allowed !