CA Mani
(Credit Analyst)
(2814 Points)
Replied 06 December 2009
I leave that chapter in my CA PE-1 exam so i cant help
CA Hemant KUMAR
(professional)
(37 Points)
Replied 07 December 2009
none of these at a maximum of 5 trials
Deep@k $ingh@l
(CA>FINAL May'14)
(595 Points)
Replied 07 December 2009
Hareesh H Sharma
(Cleared IPCC..now article)
(894 Points)
Replied 08 December 2009
Total num of chances are 10..As there are 5 bulbs wit two pints..they are
Bulb 1 paired wit bulb 2,1 with 3,1 with 4, 1& 5, 2& 3,2 &4, 2 & 5, 3 & 4, 3&5, 4& 5
tats y i said 10...i guess for permutation combination stuff there is some formulas..some 5C2 or 5P2...I went thru all this long bak...so u ppl should know it bttr
| Originally posted by :Deepak | ||
| " | I thnk answer is 7. If I m wrong, prove right answer. Ok | " |
Hi Deepak can you say the logic behind your ans. I came across this prob in scanner in that the ans. is given as 7 only
Deep@k $ingh@l
(CA>FINAL May'14)
(595 Points)
Replied 09 December 2009
Hareesh H Sharma
(Cleared IPCC..now article)
(894 Points)
Replied 09 December 2009
Deepak are u sure??Cos how do u know which is the right bulb in the 2nd point..For that only we are doing this permutation comb stuff right....I mean i studied it long back as i said so clarifying my dbt...
Hareesh H Sharma
(Cleared IPCC..now article)
(894 Points)
Replied 09 December 2009
I posted 10 after taking into account 5c2 combination
5c2 = 5!/2! *(5-2)!
=10
Hareesh H Sharma
(Cleared IPCC..now article)
(894 Points)
Replied 10 December 2009
Fine everyone i don say am right..the ans may be 7..But i was doubtful about the way he explained it...
yasmin
(law)
(23 Points)
Replied 21 December 2010
| Originally posted by : Hareesh H Sharma | ||
 |
I think its 10 |
 |
anwer is 3
Rahul Agarwal
(N.A.)
(111 Points)
Replied 10 January 2011
answer is 7 and it is right.
logic: as 3 bulbs are defective out of 5 so let me just name them as d1, d2, d3 and 2 good ones as g1 and g2.
now the room has to be lit, which can be possible even if one of the two good ones are lit, so:
d1,d2 d2,d1 d1,d3 d3,d1 d2,d3 d3,d2 when chosen the room will not lit but as soon as
d1,g1 g1,d1 d2,g1 g1,d2 d3,g1 g1,d3 d1,g2 g2,d1 d2,g2 g2d2 d3,g2 g2d3 combination is chosen the room will be lit.. so by 7th combination the room will be lit..
if this helped, "thank user"
if the doubt still persists post again.
Girish K
(Senior Accounts Executive)
(218 Points)
Replied 24 January 2011
If I m not wrong... see the answer
Total Comb is 5C2 = 5 X 4 = 20 (Here 2 is the Points)
Less: Damaged 3C2 = 3 X 2 = 6
So Ans is 14....
| Point1 | Point2 |
| d1 | d2 |
| d1 | d3 |
| d1 | g1 |
| d1 | g2 |
| d2 | d1 |
| d2 | d3 |
| d2 | g1 |
| d2 | g2 |
| d3 | d1 |
| d3 | d2 |
| d3 | g1 |
| d3 | g2 |
| g1 | d1 |
| g1 | d2 |
| g1 | d3 |
| g1 | g2 |
| g2 | d1 |
| g2 | d2 |
| g2 | d3 |
| g2 | g1 |
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