@deepkarg said:What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?17618754043748249
187
Please solve it .
@meenu05 said:Please solve it .
The given expression is equal to:
log5(3125 * 125^(1/4))/log5(5^(1/4))
=(5+0.75)/0.25=5.75/0.25=23
log(base 5^1/4)(5^5*5^(3/4))=23
@Logrhythm , @ravi.theja , @mailtoankit , @bs0409 , @legion20 ek question main bahut dimaag lag raha hai bhaiyyo .... it follows .... In how many ways can a pack of 52 cards be divided into 4 sets , 3 of them having 16 cards each and the fourth just 4 cards...? a) 16!*52! b)(52!)/(16!)^3 c) 52!/(3!)^16 d) 52!/((16!)^3 * (3!))
@uutkarshsingh said:@Logrhythm , @ravi.theja , @mailtoankit , @bs0409 , @legion20 ek question main bahut dimaag lag raha hai bhaiyyo .... it follows .... In how many ways can a pack of 52 cards be divided into 4 sets , 3 of them having 16 cards each and the fourth just 4 cards...? a) 16!*52! b)(52!)/(16!)^3 c) 52!/(3!)^16 d) 52!/((16!)^3 * (3!))
is it option d??
YESH!! it is but mera yeh nahi aa raha... mera hai (52!)/(4*(16!)^3*3)... explain kar do plz... main apna bhi approach dalunga
@uutkarshsingh said:@Logrhythm , @ravi.theja , @mailtoankit , @bs0409 , @legion20 ek question main bahut dimaag lag raha hai bhaiyyo .... it follows .... In how many ways can a pack of 52 cards be divided into 4 sets , 3 of them having 16 cards each and the fourth just 4 cards...? a) 16!*52! b)(52!)/(16!)^3 c) 52!/(3!)^16 d) 52!/((16!)^3 * (3!))
option d?
16 16 16 4 ----> 52c16*36c16*20c16*4c4*4!/3!------>52!/((16)^3*3!) ?
@uutkarshsingh said:YESH!! it is but mera yeh nahi aa raha... mera hai (52!)/(4*(16!)^3*3)... explain kar do plz... main apna bhi approach dalunga@Logrhythm
52c16*36c16*20c16*4!/3! = option d...
@TootaHuaDil @bs0409 @Logrhythm
@TootaHuaDil said:There are 100 players from 1 to 100 and 100 baskets from 1 to 100.1st player puts 1 ball in every basket starting from the first baskt , second puts 2 in each an so on but starting from the second basket ( in baskets 2 , 4 , 6 ..) Third player puts 3 in baskets in 3 , 6 , 9 .. .... Process continues till 100th player puts 100 balls in 100th basket ..Which basket wil have max no of balls(a)96 (b)98(c)100(d)none f thesePlease explain the approach too..
consider the box number 96-
factors of 96 => 1, 2,3,4,6,8,12,16,24,32,48,96
i think this means 1st player puts 1 ball in it , 2nd player puts 2 ball in it , 3rd puts 3 balls 4th puts 4 balls , 6th player puts 6 ball in that basket..... so on till the 96th player puts 96 balls..... in short , the total numbers of balls will be the sum of their factors ...
the sum of factors is max for 96 .. so 96 should be the answer ... wat do you say guys ... ??
@mailtoankit @Logrhythm
The way you people have solved is correct... but i am failing to understand why my method is wrong... ... MY APPROACH : first lets select 48 out of 52..... i.e 52c48.... then distribute 48 cards in 3 sets ... 16 in each... i.e (48)!/((16!)^3*(3!))...then multiply both... i.e 52c48 * (48)!/((16!)^3*(3!)) ....... = (52!)/(4*(16!)^3*3) .... divisor main mera extra 4 aa raha hai.. i dont knw wh its wrng ..aprch orrct lag rahi hai....
what is the ans of this question? Find the value of x^3 if x^2+x+1=0.