Permutation & Combination? Very Urgent.......

 I think its 10

I leave that chapter in my CA PE-1 exam so i cant help

none of these at a maximum of 5 trials

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

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

 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...

I posted 10 after taking into account 5c2 combination

5c2 = 5!/2! *(5-2)!

        =10

I think deepak is right...the ans should b 7 trials...

Fine everyone i don say am right..the ans may be 7..But i was doubtful about the way he explained it...

Originally posted by : Hareesh H Sharma
	I think its 10

anwer is 3

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.

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....