@catahead said:What is the ratio of the surface area of a sphere of radius 1cm and a hemisphere which has the same volume as the sphere ?
2^1/3 : 1
@catahead said:What is the ratio of the surface area of a sphere of radius 1cm and a hemisphere which has the same volume as the sphere ?
radius of the hemisphere = cube root of 2
surface area of sphere = 4*pi*r^3 = 4*pi
surface area = 3*pi*r^2 = 3*pi*(2)^2/3
ratio = 4/3*(2)^(-2/3)
@catahead said:What is the ratio of the surface area of a sphere of radius 1cm and a hemisphere which has the same volume as the sphere ?
2^4/3 : 3
How many words of 4 different letters can be formed out of 7 capital letters, 3 vowels and 5 consonants if each word starts with a capital letter and contains atleast 1 vowel?
@anuragu79 said:How many words of 4 different letters can be formed out of 7 capital letters, 3 vowels and 5 consonants if each word starts with a capital letter and contains atleast 1 vowel?
8358
@anuragu79 said:How many words of 4 different letters can be formed out of 7 capital letters, 3 vowels and 5 consonants if each word starts with a capital letter and contains atleast 1 vowel?
first place 7 ways.
now case 1 : when 1 vowel is present : 7*3*11*10 = 2310*3(as vowel can be in any of the 3 place) = 6930
case 2: when 2 vowels are present : 7*3C2*3*11 = 693
case 3 : when all vowels are presnt : 7*3! = 42
total = 6930+693+42 = 7665 😞 😞
@anuragu79 said:How many words of 4 different letters can be formed out of 7 capital letters, 3 vowels and 5 consonants if each word starts with a capital letter and contains atleast 1 vowel?
672?
@chandrakant.k said:first place 7 ways.now case 1 : when 1 vowel is present : 7*3*11*10 = 2310*3(as vowel can be in any of the 3 place) = 6930case 2: when 2 vowels are present : 7*3C2*3*11 = 693case 3 : when all vowels are presnt : 7*3! = 42total = 6930+693+42 = 7665
in case 2 shud we not multiply by 3! and not 3..........
@Dexian said:in case 2 shud we not multiply by 3! and not 3..........
Yup. Sorry answer iwould be 8358
@anuragu79 lagta hai ye boss gaya idar se.. without confirming answer 😞
@iLoveTorres method??
@bodhi_vriksha said:Decagon is a 2D figure ..it's a polygon with 10 sides.One from my side: What is the volume of the largest cone that can be inscribed in a unit sphere.Team BV
32*pi/81*R^3
radius of sphere = R
radius of cone = r
height of cone = h
r^2 + (h - R)^2 = R^2
r^2 = R^2 - (h - R)^2
V = 1/3*pi*r^2*h
= 1/3*pi*(R^2 - (h - R)^2)*h
= 1/3*pi*(2h^2R -h^3)
dV/dh = 4*pi*h*R/3 - h^2*pi = 0----> h = 4R/3
V = 1/3*pi*(R^2 - (4R/3 - R)^2)*h
V = 32*pi*R^3/81
@chandrakant.k said:Yup. Sorry answer iwould be 8358@anuragu79 lagta hai ye boss gaya idar se.. without confirming answer @iLoveTorres method??
look capital letters cant come in between so the first place can be filled in 7 ways..
the remaining three places can be filled up as
case 1) 1 vowel + 2 consonants i.e 3c1*5c2=3*10=30 they can be arranged in 3!
case 2) 2 vowels +1 consonant i.e 3c2*5c1 = 3*5=15 they can be arranged in 3!
case 3) 3 vowels = 1 way they can be arranged in 3!
So totally there are 7*((30+15+1)*6)=7*276=1932
the remaining three places can be filled up as
case 1) 1 vowel + 2 consonants i.e 3c1*5c2=3*10=30 they can be arranged in 3!
case 2) 2 vowels +1 consonant i.e 3c2*5c1 = 3*5=15 they can be arranged in 3!
case 3) 3 vowels = 1 way they can be arranged in 3!
So totally there are 7*((30+15+1)*6)=7*276=1932
@anuragu79 said:How many words of 4 different letters can be formed out of 7 capital letters, 3 vowels and 5 consonants if each word starts with a capital letter and contains atleast 1 vowel?
7C1 * 14C3 * 3! - 7C1 * 11C3 * 3! = 8358
@anuragu79 said:How many words of 4 different letters can be formed out of 7 capital letters, 3 vowels and 5 consonants if each word starts with a capital letter and contains atleast 1 vowel?
7c1*3c1*11c2*3! + 7c1*3c2*11c1*3! + 7c1*3c3*3! = 8358...
@iLoveTorres said:look capital letters cant come in between so the first place can be filled in 7 ways..the remaining three places can be filled up ascase 1) 1 vowel + 2 consonants i.e 3c1*5c2=3*10=30 they can be arranged in 3! case 2) 2 vowels +1 consonant i.e 3c2*5c1 = 3*5=15 they can be arranged in 3!case 3) 3 vowels = 1 way they can be arranged in 3!So totally there are 7*((30+15+1)*6)=7*276=1932
in each cases why have taken consonents only... after first capital letter is picked and vowel is picked then the remaining letters can be consonants or Capital letters..
PS aaj itna sannata kyun hai bhai... :)
PS aaj itna sannata kyun hai bhai... :)
@Dexian said:in each cases why have taken consonents only... after first capital letter is picked and vowel is picked then the remaining letters can be consonants or Capital letters..PS aaj itna sannata kyun hai bhai...
i have not seen a word with a capital letter in between..
hi puys plz refer to the questions and quote me..
thanks in advance
@iLoveTorres said:look capital letters cant come in between so the first place can be filled in 7 ways..
the remaining three places can be filled up as
case 1) 1 vowel + 2 consonants i.e 3c1*5c2=3*10=30 they can be arranged in 3!
case 2) 2 vowels +1 consonant i.e 3c2*5c1 = 3*5=15 they can be arranged in 3!
case 3) 3 vowels = 1 way they can be arranged in 3!
So totally there are 7*((30+15+1)*6)=7*276=1932
y?? it has to start with capital letters. That doesnt mean capital letters canot come in between
@iLoveTorres said:i have not seen a word with a capital letter in between..
lekin bhai most of the word that u make .......... will also be not in dictionary... hai na...
IMO Capitals shud be cosidered...
IMO Capitals shud be cosidered...
@iLoveTorres said:i have not seen a word with a capital letter in between..
Dont go on this notion. Question just asks the possibility. Whether that is used in reality or not in not a matter of our concern 