@Amrofa said:The probability of a number n showing in a throw of dice marked 1 to 6 isproportional to n.then the probability of the number 3 showing in throw is:1. 1/22. 1/63. 1/74. 1/21
X+2X +3X+4X+5X+6X = 1 ==> 21X=1 ==> 3x= 1/7 ")
@Amrofa said:How many five-digit numbers can be formed such that it has followingproperties :I) It has at least one zero and at most 3 zeroes.II) The non-zero digits are non-repeating.9962 17378 12570 14398 15408
It will be 15408.
For 1 zero 4*9!/5!,for 2 zero 6*9!/6! for 3 zero 4*9!/7! Add them
@Amrofa said:How many five-digit numbers can be formed such that it has followingproperties :I) It has at least one zero and at most 3 zeroes.II) The non-zero digits are non-repeating.9962 17378 12570 14398 15408
15408
1. one zero-
_ _ _ _ _
The zero can occupy only one of the last four places.
Selecting the place in 4C1 ways.
remaining 4 places can be filled by 9*8*7*6
therefore, 4*9*8*7*6
2. 2 zeroes-
_ _ _ _ _
The zeroes can occupy only two of the last four places.
Selecting the place in 4C2 ways.
remaining 3 places can be filled by 9*8*7
therefore, 6*9*8*7
3. 3 zeroes-
_ _ _ _ _
The zeroes can occupy only three of the last four places.
Selecting the place in 4C3 ways.
remaining 2 places can be filled by 9*8
therefore, 4*9*8
total = 72( 28*6+ 42+ 4)
72( 172+42)
72(214)
=15408
1. one zero-
_ _ _ _ _
The zero can occupy only one of the last four places.
Selecting the place in 4C1 ways.
remaining 4 places can be filled by 9*8*7*6
therefore, 4*9*8*7*6
2. 2 zeroes-
_ _ _ _ _
The zeroes can occupy only two of the last four places.
Selecting the place in 4C2 ways.
remaining 3 places can be filled by 9*8*7
therefore, 6*9*8*7
3. 3 zeroes-
_ _ _ _ _
The zeroes can occupy only three of the last four places.
Selecting the place in 4C3 ways.
remaining 2 places can be filled by 9*8
therefore, 4*9*8
total = 72( 28*6+ 42+ 4)
72( 172+42)
72(214)
=15408
@Amrofa said:How many five-digit numbers can be formed such that it has followingproperties :I) It has at least one zero and at most 3 zeroes.II) The non-zero digits are non-repeating.9962 17378 12570 14398 15408
4c1*9*8*7*6 + 4c2*9*8*7+ 4c3*9*8 = 15408
@vbhvgupta said:how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...1. 3 terms2. 4 terms3. 5 terms4. 6 termsCan someone explain me the concept behind it????
It depends on how many terms you are taking for atleast 3,4,5,6....becoz common difference you are taking changes everytime. Use Last term =first term+ (n-1)diff and check
@vbhvgupta said:how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...1. 3 terms2. 4 terms3. 5 terms4. 6 termsCan someone explain me the concept behind it????
take the difference and find the number of factors 3535-1235 =2300
2300=12*5*5*2*2 =23*5^2 *2^2
Total Number of factors =2*3*3=18
1) Atleast 3 terms = 18-1 =17 (we wil take all the factors except 2300 because with the difference of 2300 it will make AP of only 2 terms i.e 1235,3535)
2)Atleast 4 terms = 18-2=16 factors ( we will not take 2300/1 and 2300/2 as difference )
3) Atleast 5 terms=18-2=16 factors (Same as above 2300/1 and 2300/2 will not consider as there is not any factor as 2300/3)
4)Atleast 6 terms =18-3=15 factors (we will not take 2300/1,2300/2,2300/4 as differ. of AP)
2300=12*5*5*2*2 =23*5^2 *2^2
Total Number of factors =2*3*3=18
1) Atleast 3 terms = 18-1 =17 (we wil take all the factors except 2300 because with the difference of 2300 it will make AP of only 2 terms i.e 1235,3535)
2)Atleast 4 terms = 18-2=16 factors ( we will not take 2300/1 and 2300/2 as difference )
3) Atleast 5 terms=18-2=16 factors (Same as above 2300/1 and 2300/2 will not consider as there is not any factor as 2300/3)
4)Atleast 6 terms =18-3=15 factors (we will not take 2300/1,2300/2,2300/4 as differ. of AP)
If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
Q.
HCF of two numbers is 99 and their LCM is 2772. The numbers are:
A. 198,1386
B. 297,924
C. 396,693
D. 693,11088
After long calculations I got the answer but it took me too much time.
So, I would know if someone here knows the shorter logical approach? (like picking answer directly from the options or something like that)
Regards.
Q.
HCF of two numbers is 99 and their LCM is 2772. The numbers are:
A. 198,1386
B. 297,924
C. 396,693
D. 693,11088
After long calculations I got the answer but it took me too much time.
So, I would know if someone here knows the shorter logical approach? (like picking answer directly from the options or something like that)
Regards.
@vbhvgupta said:how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...1. 3 terms2. 4 terms3. 5 terms4. 6 termsCan someone explain me the concept behind it????
okay,
3535= 1235+ (n-1)d
2300=(n-1)d
factors of 2300= 1, 2, 4, 5, 10, 20, 23, 25, 46, 50, 92, 115, 230, 460,575, 1150, 2300
1. n>=3
n-1 can have values from 2 to 2300
therefore 17 cases
2. n>=4
n-1>=3....16 cases
3. n>=5
n-1>=4........16 cases
4. n-1>=5 (as n>=6)
15 cases
3535= 1235+ (n-1)d
2300=(n-1)d
factors of 2300= 1, 2, 4, 5, 10, 20, 23, 25, 46, 50, 92, 115, 230, 460,575, 1150, 2300
1. n>=3
n-1 can have values from 2 to 2300
therefore 17 cases
2. n>=4
n-1>=3....16 cases
3. n>=5
n-1>=4........16 cases
4. n-1>=5 (as n>=6)
15 cases
@anytomdickandhary Sir, please help me out to solve this question using equations:
3 different dies were rolled. find the no. of ways the summation of no will be 12?
my approach:
x+y+z=12
(6-x')+(6-y')+(6-z')=12
x'+y'+z'=6
no of non negative values: 8C2=28 
where am i going wrong with the concept sir? :(
@vbhvgupta said:how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast...1. 3 terms2. 4 terms3. 5 terms4. 6 termsCan someone explain me the concept behind it????
3535-1235=2300=2^2 * 5^2 * 23
well there can be infinite terms as such. But if we take only integral values of common difference,
1. atleast 3 terms- there are 17 APs.
2. atleast 4 terms- 16 APs
3. atleast 5 terms- 16 APs
4. atleast 6 terms- 15 APs
3535-1235=2300=2^2 * 5^2 * 23
well there can be infinite terms as such. But if we take only integral values of common difference,
1. atleast 3 terms- there are 17 APs.
2. atleast 4 terms- 16 APs
3. atleast 5 terms- 16 APs
4. atleast 6 terms- 15 APs
@cadmium said:Q. HCF of two numbers is 99 and their LCM is 2772. The numbers are:A. 198,1386B. 297,924C. 396,693D. 693,11088After long calculations I got the answer but it took me too much time.So, I would know if someone here knows the shorter logical approach? (like picking answer directly from the options or something like that)Regards.
HCF is 99 let the two numbers be 99*a ,99*b
we know product of two numbers = lcm * hcf ==> 99*a*99*b = 99*2772 ==> a*b=28 ( 1,28 ) (2,14 ) , (4,7 ) ..now sure a,b are co-primes to each other ...so a=4 b=7 ==> 396,693 option C
we know product of two numbers = lcm * hcf ==> 99*a*99*b = 99*2772 ==> a*b=28 ( 1,28 ) (2,14 ) , (4,7 ) ..now sure a,b are co-primes to each other ...so a=4 b=7 ==> 396,693 option C
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
total no of possible solutions=38C2
out of which in one case x=y.
so solns in which x>y will be half of (38C2 - 1). will be 351
out of which in one case x=y.
so solns in which x>y will be half of (38C2 - 1). will be 351
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
190?
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
331??
@ravi.theja said:HCF is 99 let the two numbers be 99*a ,99*bwe know product of two numbers = lcm * hcf ==> 99*a*99*b = 99*2772 ==> a*b=28 ( 1,28 ) (2,14 ) , (4,7 ) ..now sure a,b are co-primes to each other ...so a=4 b=7 ==> 396,693 option C
Answer is Option A.
@x2maverickc said:total no of possible solutions=38C2out of which in one case x=y.so solns in which x>y will be half of (38C2 - 1). will be 351
@Zedai said:190?
@ravi.theja said:331??
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