Latest MBA Aptitude Question SOLUTION: A box contains 20 tickets of identical appearance, the ticket being numbered 1, 2 , 3.....20. In how many ways can 3 tickets be chosen such that the numbers on the drawn tickets ar

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(#M40016780) MBA Numerical Ability question permutations and combinations, probability Keep an EYE

A box contains 20 tickets of identical appearance, the ticket being numbered 1, 2 , 3.....20. In how many ways can 3 tickets be chosen such that the numbers on the drawn tickets are in arithmetic progression?If sumone cud post the approach it wud be gr8

1) 18
2) 33
3) 56
4) 90
5) 84

Asked In MBA (12 years ago)

Unsolved Read Solution (6)

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Narendra Dhaker  (9 years ago)

Ans is 90
there could be maximum Common difference will be 9 so we have to check no of ways of AP with Common difference form 1 to 9

with C.D. 1 : 18 ways
with C.D. 1 : 16 ways
with C.D. 1 : 14 ways
with C.D. 1 : 12 ways
with C.D. 1 : 10 ways
with C.D. 1 : 8 ways
with C.D. 1 : 6 ways
with C.D. 1 : 4 ways
with C.D. 1 : 2 ways

TOtal No. of ways are 2 + 4 + 6 + 8 + 10 + 12 + 14 +16 + 18
with is 90

Is this solution Helpfull? Yes (5) | No (4)  

Atanu Roy  (10 years ago)

2+4+6+..+18=90

Is this solution Helpfull? Yes (4) | No (6)  

prasanthi  (10 years ago)

10c2*2!=90

Is this solution Helpfull? Yes (4) | No (4)  

Macha Devendher  (10 years ago)

A sequence of 3 tickets, starting at "a" and with a difference d, gives
a, (d + a), (2d + a)

For a given d, put a = 1, then final ticket number is
2d + 1

There are 20 - (2d + 1) ticket numbers above this final ticket
Thus, keeping the same d, we can increment the start point
20 - (2d + 1) times and stay in range
[ie slide the whole sequence along]

So for any given d there are
1 + 20 - (2d +1)
= 20 - 2d
= 2(10 - d)
possible ways to make an AP, and d = 1 to 9

So adding up for all d, we just get twice the sum of 1 to 9
ie sum of an AP

2{ (1/2)(9 x 10) } = 90

Is this solution Helpfull? Yes (3) | No (6)  

Nithin  (7 years ago)

consider intial a,b,c as progession
a=1,then b=x,c=2x-1.....beacuse c=a/2+b/2...then c is odd ...max of c=19,then x=10........min to max values of for a=1 are 2 to 10......9 ways......
a=2, x varies from 3 to 11....ways = 11-3+1=9..
a=3,ways =4 to 11...ways=8
a=4, ways =8.......
u will a similarity such that max a=18..
similarity= a=1,2...ways =9+9=9*2
a=2,3..ways=8*2.......son on
at a=17,18..ways=1+1=1*2
..............total no of ways=2*(1+2+3+4+5+6+7+8+9)=45*2=90
option D=90

Is this solution Helpfull? Yes | No (2)  

KARUKU AJAY KUMAR  (6 years ago)

Common Difference(CD)-1 : (1,2,3),(2,3,4),(3,4,5),.............,(1... sets
CD-2 : (1,3,5),(2,4,6),(3,5,7),.........,(16,18... sets
CD-3 : (1,4,7),(2,5,8),(3,6,9),........,(14,17,... sets
and so on upto CD-8
for CD-9 there are only two sets i.e., (1,10,19),(2,11,20)
therefore,sum of all sets =18+16+14+............+4+2=90

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