had edited my postthought process: let ABCD be the four numbers chosencase 1) let A and B be the same, ways to selecting A/B = 6ways of selecting C = 5ways of selecting D = 4
What if B and C are same instead? We need to also choose which 2 are same (4C2 = 6 ways)
case 4) A and B are same and C and D are same, ways of selecting A/B = 6ways of selecting C/D = 5
As above need to choose which two pairs are same in 3 ways so 6 * 5 * 3 = 90 ways
Total = 710 + 120 + 6 + 90 = 936/1296 = 13 / 18.
Answer to same aayega. But this has entirely too much probability (pun intended!) of going wrong (I messed up the calcs once while writing this post!)...I prefer what I did earlier 😃 One step and poof... regards scrabbler
What if B and C are same instead? We need to also choose which 2 are same (4C2 = 6 ways)so 6 * 5 * 4 * 6 = 720As above need to choose which 3 are same (4 ways) so 6 * 5 * 4 = 120 waysThis no change!As above need to choose which two pairs are same in 3 ways so 6 * 5 * 3 = 90 waysTotal = 710 + 120 + 6 + 90 = 936/1296 = 13 / 18.Answer to same aayega. But this has entirely too much probability (pun intended!) of going wrong (I messed up the calcs once while writing this post!)...I prefer what I did earlier One step and poof...regardsscrabbler
sir OA states it's 11/46. Thoda explanation likhiyae. it will be useful for us, since the question is little tricky.
Oops, yeah, 1 - 35/46 karne bhool gaya :(
Hold on to your seats, the explanation is long! I did with fig which made it short to write...but to explain the thoughts requires too many words it seems :(
So I calculated the prob that there are no 2 next to each other.
Let the guys in order b A B C D till Y. I started off assuming one guy picked, let's say A. So then B and Y cannot be picked. Now any of the other 22 can be picked second.
But in the case of C being picked there are only three others besides A and C who are eliminated for the third pick B, Y and D and so 20 ways to pick the third, similarly if X is the second one then B, Y and W are out so 20 ways.
If any of the other 20 from D to W are picked second, there are 4 people eliminated for the third pick (suppose F is picked for example, A's neighbours B and Y and that F's neighbours E and G) and so there are only 19 ways to pick the 3rd.
So total ways to pick 2nd and 3rd in order is (2 * 20 + 20 *19) = 420
Same would hold for any of the 25 being picked 1st. So the number of ways of picking the three in order is 25 * 420
Now if no condition is there, number of ways of picking 3 in order is 25P 3 = 25 * 24 * 23. So prob that no two are neighbours is 25 * 420 / 25 * 24 * 23 = 35 / 46. Hence at least one neighbouring pair will happen in 1 - 35/46 = 11 / 46 ways. hence 57. regards scrabbler
Hold on to your seats, the explanation is long! I did with fig which made it short to write...but to explain the thoughts requires too many words it seems
So I calculated the prob that there are no 2 next to each other.
Let the guys in order b A B C D till Y. I started off assuming one guy picked, let's say A. So then B and Y cannot be picked. Now any of the other 22 can be picked second.
But in the case of C being picked there are only three others besides A and C who are eliminated for the third pick B, Y and D and so 20 ways to pick the third, similarly if X is the second one then B, Y and W are out so 20 ways.
If any of the other 20 from D to W are picked second, there are 4 people eliminated for the third pick (suppose F is picked for example, A's neighbours B and Y and that F's neighbours E and G) and so there are only 19 ways to pick the 3rd.
So total ways to pick 2nd and 3rd in order is (2 * 20 + 20 *19) = 420
Same would hold for any of the 25 being picked 1st. So the number of ways of picking the three in order is 25 * 420
Now if no condition is there, number of ways of pickling 3 in order is 25P 3 = 25 * 24 * 23. So prob that no two are neighbours is 25 * 420 / 25 * 24 * 23 = 35 / 46. Hence at least one neighbouring pair will happen in 1 - 35/46 = 11 / 46 ways. hence 57.
regardsscrabbler
there is an easier way to do it:
let 2 ppl sit together:
now 25 such pairs exist.. 25C1.
so wen any pairs gets picked 2 more around them are disqualified for 3rd pick...
there is an easier way to do it:let 2 ppl sit together:now 25 such pairs exist.. 25C1.so wen any pairs gets picked 2 more around them are disqualified for 3rd pick...so total 25 - 4 =21 for 3rd pick..total 25C1 * 21case II: let all 3 sit together:25C1so P= 550/25C3=11/46
