Chk attachment for ques
@pankaj1988 said:Chk attachment for ques
5^(a2-1)=1/(1+2/3)+1=1/(5/3)+1=3/5+1=8/5
5^a2=8
5^a3/8=11/8
5^a3=11
5^a4=14
..
so ak=8+(k-2)*3
for k=41 , ak=125
but 5^3=125
so a41=3
41
5^a2=8
5^a3/8=11/8
5^a3=11
5^a4=14
..
so ak=8+(k-2)*3
for k=41 , ak=125
but 5^3=125
so a41=3
41
@jain4444 said:In how many ways can 12 people be arranged around a hexagonal table with successive sides being of length a metres, b metres, c metres, a metres, b metres and c metres respectively, if people can sit on chairs placed at corners and the centres of sides?1) 36 — 12!/2^62) 6! — 12!/2^63) 12!/2^64) None of these
im getting 12!/2
@jain4444 said:In how many ways can 12 people be arranged around a hexagonal table with successive sides being of length a metres, b metres, c metres, a metres, b metres and c metres respectively, if people can sit on chairs placed at corners and the centres of sides?1) 36 — 12!/2^62) 6! — 12!/2^63) 12!/2^64) None of these
3) 12!/2^6 ?? 
@jain4444 said:In how many ways can 12 people be arranged around a hexagonal table with successive sides being of length a metres, b metres, c metres, a metres, b metres and c metres respectively, if people can sit on chairs placed at corners and the centres of sides?1) 36 — 12!/2^62) 6! — 12!/2^63) 12!/2^64) None of these
12!/2 for me too
@krum said:im getting 12!/2
@ScareCrow28 said:3) 12!/2^6 ??
approach daloo bhai log yahan boht ajeeb sa aa raha hai
@chillfactor sir jii madad
@jain4444 said:approach daloo bhai log yahan boht ajeeb sa aa raha hai @chillfactor sir jii madad
if it were normal circle - 11!
but here each person has the option to choose from 6 seats (center-a,b,c; corner-a,b,c)
so 11! * 6 = 12! / 2
but here each person has the option to choose from 6 seats (center-a,b,c; corner-a,b,c)
so 11! * 6 = 12! / 2
@jain4444 said:approach daloo bhai log yahan boht ajeeb sa aa raha hai @chillfactor sir jii madad
Sirjee I did like this:
12 places..Cut the hexagon in two halves..they will be alike to each-other (But sides are different)
Now, There are 12 places for 1st person, but since there are 2 alike places..divide by 2
So, (12/2)*(11/2)*(10/2)*(9/2)*(8/2)*(7/2)
Now 6 places left so 6! ways
Total = 12!/2^6??
@jain4444 said:approach daloo bhai log yahan boht ajeeb sa aa raha hai @chillfactor sir jii madad
12 distinct positions hai table mai. The only line of symmetry being ki opposite sides equal hai, to basically necklace jaisa ho gaya, ulta karenge to same arrangement aa jayega, to divide by 2.
@ScareCrow28 said:Sirjee I did like this:12 places..Cut the hexagon in two halves..they will be alike to each-other (But sides are different)Now, There are 12 places for 1st person, but since there are 2 alike places..divide by 2So, (12/2)*(11/2)*(10/2)*(9/2)*(8/2)*(7/2)Now 6 places left so 6! waysTotal = 12!/2^6??
When we make the second person sit, why do we divide by two, his relative positioning with the first person will also matter?
@pankaj1988 said:Chk attachment for ques
5^a1, 5^a2, 5^a3.... making an A.P.
common difference =3.
so we have to look for the term 25, 125, or 625 and so on in the series.
125 satisfies for a=41.
P.S. Which camera have you used? brightness just blinded me for a sec :P
@ankita14 said:When we make the second person sit, why do we divide by two, his relative positioning with the first person will also matter?
We divide by two, because like the first person, he also have two alike places and similarly for others too. I don't get what's relative position has to do with not dividing by two!
@jain4444 : 6*11!
The first person can choose to be seated in 6 ways - left corner of side a, middle seat of side a, right seat of side a and so on till the middle seat of side c. Rest people can be arranged in 11! ways . So , 6*11!
@jain4444 said:In how many ways can 12 people be arranged around a hexagonal table with successive sides being of length a metres, b metres, c metres, a metres, b metres and c metres respectively, if people can sit on chairs placed at corners and the centres of sides?1) 36 — 12!/2^62) 6! — 12!/2^63) 12!/2^64) None of these
12/2*11!=12/2!
so, none of these?
@ScareCrow28 said:We divide by two, because like the first person, he also have two alike places and similarly for others too. I don't get what's relative position has to do with not dividing by two!
Say you've already seated person A. Now it will matter whether you place person B next to him or on the equivalent seat in the opposite side. They are two distinct arrangements.
@ScareCrow28 said:We divide by two, because like the first person, he also have two alike places and similarly for others too. I don't get what's relative position has to do with not dividing by two!
Yar i have a doubt on it. when first person has already started, then for the second person 11 place left. One place is fixed. So, there is no symmetry left now. a1b1xxxxxxxxxx and a1xxxxxxxb2xxx both are different cases.
Guys...koi xat qa ka thread banao. shuru karte hai. xat ki qa is two notches above the normal qa that we practice.
@jain4444 @vijay_chandola - karo bhailog start. mein office se join karta hun. (tag kar dena) \m/
@Enceladus said:Guys...koi xat qa ka thread banao. shuru karte hai. xat ki qa is two notches above the normal qa that we practice.@jain4444@vijay_chandola - karo bhailog start. mein office se join karta hun. (tag kar dena) \m/
@Enceladus said:Guys...koi xat qa ka thread banao. shuru karte hai. xat ki qa is two notches above the normal qa that we practice.@jain4444@vijay_chandola - karo bhailog start. mein office se join karta hun. (tag kar dena) \m/
leyo sood sir xat question 
From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then
they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 1. The counting would be done in a clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore the first person to be eliminated would be
the contender numbered 2.
23. Which position should a contender choose if he has to be the leader? (5 marks)
(1) 3 (2) 67 (3) 195 (4) 323 (5) 451
they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 1. The counting would be done in a clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore the first person to be eliminated would be
the contender numbered 2.
23. Which position should a contender choose if he has to be the leader? (5 marks)
(1) 3 (2) 67 (3) 195 (4) 323 (5) 451