Qualification > Math
S1 and S2 DOUBTS HERE!!!!
perish007:
this question is of statistics S1
Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
Tohru Kyo Sohma:
--- Quote from: perish007 on August 19, 2011, 11:42:38 am ---this question is of statistics S1
Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the digits is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
--- End quote ---
im not sure but is it like this
(a)6!/3!
(b)i didnt get the qns
i dont remember much but if u have the answer u can check with it
astarmathsandphysics:
Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
Must have 2 4's, or 2 5's or both.
If 2 4's there are 6!/2! different numbers and same for two 5's.
If 2 4's and two 5's then there are 7!/2!2! ways for each choice of the other two numbers. There are three ways the other two can be chosen so altogether 2*6!/2! +3*6!/2! ways altogether.
a) if the number is odd then the last digit is 3, 5 or 7.
If the last digit is 3, then the other 5 numbers can be chosen from 4,4,5,5,6,7
Following the logic above there are 2*5!/2! +2*5!/2! ways of choosing these five. The same result is is 7 is the last number.
If 5 is the last number, and the other 5 numbers contain one 4, there are 5! ways and 5!/2! ways if two 4's.
Hence preb of being odd is (2(2*5!/2! +2*5!/2!)+5!+5!/2!)/2*6!/2! +3*6!/2!)
b)Two 5's so one 4. There are 6!/2! ways this can be down. Prob = 6!/2!/(Find the total number of different 6-digits numbers that can be formed by using the digits 3, 4, 4, 5, 5, 6, 7 .
if one of the number is picked at random, find the probability that
(a) the number is odd,
(b) the number has two digits 5.
Must have 2 4's, or 2 5's or both.
If 2 4's there are 6!/2! different numbers and same for two 5's.
If 2 4's and two 5's then there are 6!/2!2! ways for each choice of the other two numbers. There are three ways the other two can be chosen so altogether 2*6!/2! +3*6!/2! ways altogether.
a) if the number is odd then the last digit is 3, 5 or 7.
If the last digit is 3, then the other 5 numbers can be chosen from 4,4,5,5,6,7
Following the logic above there are 2*5!/2! +2*5!/2! ways of choosing these five. The same result is is 7 is the last number.
If 5 is the last number, and the other 5 numbers contain one 4, there are 5! ways and 5!/2! ways if two 4's.
Hence preb of being odd is (2(2*5!/2! +2*5!/2!)+5!+5!/2!)/2*6!/2! +3*6!/2!)
b)Two 5's so one 4. There are 6!/2! ways this can be done so 6!/2!/(2*6!/2! +3*6!/2!)
ashwinkandel:
Eight People sit in a minibus: four on the sunny side and four on the shady side. If two people want to sit on opposite sides to each other and another two people want to sit on the shady sides, in how many ways can this be done?
astarmathsandphysics:
here
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