@saurav5517 said:Can Fibonacci series be applied in this question?
Perhaps....can't see a way to do so though!
regards
scrabbler
regards
scrabbler
@ravi6389 said:hi Can some one tell me whats wrong in my approach to the following question??Q. 4 dice are thrown.In how many ways can a sum of 20 be obtained?MY APPROACH ->we have a + b +c + d =20we have 2 conditions to be taken care of :->a,b, c and d >0 .....1>and a ,b,c,d So starting by 1>we have a'+1 +b' +1 +c' +1+d' +1 =20=>a'+b'+c'+d'=16So total solutions 19c3.Now for 2nd partlet us assume a' = A+6So, A + 6+b' + c'+d' =16Then,A+b'+c'+d' = 10So number of solutions = 13 c 3.Now we can select any of A, b',c' and d' to be 6So final ans -> 19 c 3 - 4 *13 c 3But Problem is answer comes out to be -ve.Any help pointing out my mistake will be appreciated
write it like this
a+b+c+d=20
as a,b,c,d has to be less or equal to 6
so 6-a +6-b +6-c +6-d=20
a+b+c+d=4
solutions 7c3=35 ways
1) A class of seven students is doing a Secret Santa, for which all seven students have contributed a gift each. However, one of the seven students has not arrived yet. The teacher decides to randomly assign each of the six students present to a gift that is not their own. If P is the probability that the seventh student is left with his own gift once he arrives, fifnd P.?
@albiesriram said:1) A class of seven students is doing a Secret Santa, for which all seven students have contributed a gift each. However, one of the seven students has not arrived yet. The teacher decides to randomly assign each of the six students present to a gift that is not their own. If P is the probability that the seventh student is left with his own gift once he arrives, fifnd P.?
The 6 kids present in the class never get their own gifts. So there are 2 cases :
a) The 7th student's gift remains untouched -> Derangement of 6 gifts
= 6!( 1/2! - 1/3! + 1/4! - 1/5! + 1/6!) = 265 = K
b) The 7th student's gift ends up with one of the 6 students -> Derangement of 7 gifts
= 7!( 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - 1/7!) = 7*K - 1
= 7!( 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - 1/7!) = 7*K - 1
Desired case is case a)
Prob = K/8K-1 = 265/2119
how to do this type of questions
Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100!
Number of digits in the number €˜a €™ are
Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100!
Number of digits in the number €˜a €™ are
@fireatwill said:how to do this type of questions Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100!Number of digits in the number €˜a €™ are
you dont do em. you skip em 
@fireatwill said:@Angadbir dada but is there any method .
there is, but it gets quite convoluted and I can assure you such convoluted questions would not make place into the CAT paper.
You can see that 100! has most no. of digits. it has 2 more digits than 99!. 99! has 2 more digits than 98!
You can safely conclude that total no. of digits in the sum is total no. of digits in 100!
Now you'd have to work with logs. log of a number to base 10 tells you about no. of digits in that no. (need to add 1 to the result)
eg, log(100) = 2 -> no. of digits = 2+1 = 3
log(1000) = 3 -> no. of digits = 3 + 1 = 4
[Log(100!)] +1 = no. of digits in 100! (notice the greatest integer function [])
Here you'd have to represent 100! as product of prime powers and add all the log results.
By the time I come to this point of realization, I'll skip the question.
There might be derived formulas for such problems, but again - they would fall beyond the purview of CAT.
@fireatwill said:how to do this type of questions Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100!Number of digits in the number €˜a €™ are
i am getting 158
number of digits of any positive integer x is exactly equal to plus 1 ie log100 =2 so no. of digits will be 2 + 1 =3
X! = x^x *e^ (-x) sqrt(2 *pi * x)
100! = 100^100 * e^(-100) sqrt(2*pi*100)
log 100! = (100 log 100) - (100 log e) +
which is equals to
= 157
So number of digits = 157+1 = 158
100! = 100^100 * e^(-100) sqrt(2*pi*100)
log 100! = (100 log 100) - (100 log e) +
which is equals to
= 157
So number of digits = 157+1 = 158
(If i am wrong then correct me)
@shattereddream said:i am getting 158 number of digits of any positive integer x is exactly equal to plus 1 ie log100 =2 so no. of digits will be 2 + 1 =3 X! = x^x *e^ (-x) sqrt(2 *pi * x) 100! = 100^100 * e^(-100) sqrt(2*pi*100)log 100! = (100 log 100) - (100 log e) +which is equals to = 157So number of digits = 157+1 = 158 (If i am wrong then correct me)
ish correct!
128^1000Mod153 = ?
@shattereddream bro can u explain the last three steps
X! = x^x *e^ (-x) sqrt(2 *pi * x)
100! = 100^100 * e^(-100) sqrt(2*pi*100)
log 100! = (100 log 100) - (100 log e) +
@albiesriram said:1) A class of seven students is doing a Secret Santa, for which all seven students have contributed a gift each. However, one of the seven students has not arrived yet. The teacher decides to randomly assign each of the six students present to a gift that is not their own. If P is the probability that the seventh student is left with his own gift once he arrives, fifnd P.?
6!/6! *8 -1 = 0.125 ??
In a triangle ABC,side BC measures 20cm and the length of the median to side BC is equal to half of the
length of the side BC.What is the ratio of the sum of the
squares of the sides of the triangle ABC to the sum of the squares of its medians ?
length of the side BC.What is the ratio of the sum of the
squares of the sides of the triangle ABC to the sum of the squares of its medians ?
@nole
well theres a formula that 3*sum of squares of sides=4* SUM OF SQUARES OF MEDIANS. the value of cb is unecessary, you can derive the formula by using apoolonius for all three sides
How many zeroes are there in 100! Using the base 5