Q1. 7*7 *4 = 49*4 = 196Originally Posted by awais
(if no repetition, then 6*5 *4 =30*4 = 120)
Q2.
6561*9
729*7
81*5
9*3
Total=64584
Total 5digit numbers=90000
P = 64584/90000 ~ 0.71
Q1. 7*7 *4 = 49*4 = 196Originally Posted by awais
(if no repetition, then 6*5 *4 =30*4 = 120)
Q2.
6561*9
729*7
81*5
9*3
Total=64584
Total 5digit numbers=90000
P = 64584/90000 ~ 0.71
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make way for th bull -- Thebullfighter.
Happy New Year 2005 to All.
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Q1
a) with repetition
7*7*4=196
b)without repetition
first two digits -both even 3!/1!=6 times number of odd digits left *4 =24
first two digits -both odd 4!/2!=12 times number of odd digits left *2 =24
first two digits -one even one odd 3*4=12 times number of odd digits left *3=36
asw =84
Q2.
in order to have product of any number of numbers even, one of numbers must be even.
The probability that all five digits are odd (assuming that 0 is even) is:
(5/10)^5 = (1/2)^5 =1/32 the probability that at least one of digits is even =31/32
asw 31/32
yap. second is 31/32
just learned th method segregation from another thread.
Opportunities seem Bigger going, than coming.
make way for th bull -- Thebullfighter.
Happy New Year 2005 to All.
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u testin' me or u din't actually get it... gotta be kiddin me, i learned P&C & P analysin thy answers & studyin links provided by u..Originally Posted by shaq
Q.1. How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6 and 7?
odd no.s-- 1,3,5,7
Case 1-- Last digit is 1 (fixed)
so other positions can be filled in (7-1=6): 6*5 ways
similarly for other 3 odd numbers.
Hence, 6*5 *4=120
how did i do???
Opportunities seem Bigger going, than coming.
make way for th bull -- Thebullfighter.
Happy New Year 2005 to All.
AID Appeal for Tsunami Disaster Relief
heeey... thanx man. (shaq just praised mine over 'shaq's' own method, tht too for P.. now, this is an honour.)
Opportunities seem Bigger going, than coming.
make way for th bull -- Thebullfighter.
Happy New Year 2005 to All.
AID Appeal for Tsunami Disaster Relief
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