@bodhi_vriksha said:Ext. Qs. A cubical die is thrown and the outcomes p, q, r, s, t are noted what is the chance that p ≤q ≤r ≤s ≤ t?Team BV - Kamal Lohia
oh did not read "die"
would be 10c5
@bodhi_vriksha said:Ext. Qs. A cubical die is thrown and the outcomes p, q, r, s, t are noted what is the chance that p ≤q ≤r ≤s ≤ t?Team BV - Kamal Lohia
p, Q = q+1,R=r+2,S =s+3,T=t+4 and range is 1 to 10 so, 10C5 ?? 
@bodhi_vriksha said:Ext. Qs. A cubical die is thrown and the outcomes p, q, r, s, t are noted what is the chance that p ≤q ≤r ≤s ≤ t?Team BV - Kamal Lohia
@pankaj1988 said:An intelligence agency forms two digit code consisting of distinct digits selected from 0 through 9 such that first digit is not 0. Some codes, when hand written on slip, can however potentially create confusion when read upside down- for example 61 may appear as 19. Find the number of codes for which there is no confusion.a)73 b)70 c)71 d)67No OA
71...
@Logrhythm said:13c5
Dice mein max number 6 hota hai..
So, min 1; max 6+4 = 10.. Hence, 10C5 I suppose..
So, min 1; max 6+4 = 10.. Hence, 10C5 I suppose..
@bodhi_vriksha said:Ext. Qs. A cubical die is thrown and the outcomes p, q, r, s, t are noted what is the chance that p ≤q ≤r ≤s ≤ t?Team BV - Kamal Lohia
6C5 + 6C4*2 + 6C3*3 + 6C2*4 + 6C1 + 6C2*2 + 6C3*3 = 252
One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide as shown in the figure given below. What is the total surface area of the solid thus formed?
a)88 pi b)83 pi c)75 pi d)73 pi
a)88 pi b)83 pi c)75 pi d)73 pi
@bodhi_vriksha said:Let me take some example:
Find the number of four digit numbers abcd such that a
Clearly a, b, c, d aredistinct positive integersvarying from 1 to 9.
So they can be selected in C(9, 4) ways and arranged in one possible arrangement only. Thus required numbers are C(9, 4)
Next now..Find the number of four digit numbers abcd such that a ≤b ≤c ≤d.
Now the a, b, c, d can be equal also. So to remove the equality part see the trick carefully.
We know that a ≤ b, can you tell me smallest number which is certainly greater than a.
........
........
It is b + 1, I don't know whether you guessed it correctly or not.
Anyway now I can write that a ≤ b is equivalent to a
Similarly, extending this, the given inequality can be re-written as a and d + 3 can be at most 9 + 3 = 12.
And you just need to select fourdistinct positive integersfrom 1 to 12.
And this can be done in C(12, 4) ways.
Team BV - Kamal Lohiasorry sir ........... this bold part is not very clear.. can u plzz..(and d + 3 can be at most 9 + 3 = 12.
And you just need to select fourdistinct positive integersfrom 1 to 12.)
@Dexian said:sorry sir ........... this bold part is not very clear.. can u plzz..(and d + 3 can be at most 9 + 3 = 12.And you just need to select four distinct positive integersfrom 1 to 12.)
Though there is nothing like bold part in your post but your query is understandable.
See when I say a, b + 1, c + 2, d + 3 are four distinct positive integers from 1 to 12 such that a
- then the largest selection of four integers will be {9,10,11,12} for {a, b+1, c+2, d+3} which corresponds to {a,b,c,d} = {9,9,9,9} i.e. 9999 which is largest permissible number
- and the smallest selection of four integers will be {1,2,3,4} for {a, b+1, c+2, d+3} which corresponds to {a,b,c,d} = {1,1,1,1} i.e. 1111 which is smallest permissible number.
Does this make sense to you? :)
Team BV - Kamal Lohia
See when I say a, b + 1, c + 2, d + 3 are four distinct positive integers from 1 to 12 such that a
- then the largest selection of four integers will be {9,10,11,12} for {a, b+1, c+2, d+3} which corresponds to {a,b,c,d} = {9,9,9,9} i.e. 9999 which is largest permissible number
- and the smallest selection of four integers will be {1,2,3,4} for {a, b+1, c+2, d+3} which corresponds to {a,b,c,d} = {1,1,1,1} i.e. 1111 which is smallest permissible number.
Does this make sense to you? :)
Team BV - Kamal Lohia
@Koushik98 said:One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide as shown in the figure given below. What is the total surface area of the solid thus formed? a)88 pi b)83 pi c)75 pi d)73 pi
It'll be simply 1/4(4*pi*6^2 + 4*pi*4^2) + pi*6^2 = 88pi
Team BV - Kamal Lohia
Team BV - Kamal Lohia
@Koushik98 said:One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide as shown in the figure given below. What is the total surface area of the solid thus formed? a)88 pi b)83 pi c)75 pi d)73 pi
a)88 pi ?
Outer surface area = 52 pi
flat bottom face surface area = 26 pi
Annular surface area = 10 pi
@bodhi_vriksha said:It'll be simply 1/4(4*pi*6^2 + 4*pi*4^2) + pi*6^2 = 88piTeam BV - Kamal Lohia
why are we not taking plane surface area of smaller sphere???
@RDN said:a)88 pi ?Outer surface area = 52 piflat bottom face surface area = 26 piAnnular surface area = 10 pi
yea 88 pi is the OA
@Koushik98 said:why are we not taking plane surface area of smaller sphere???
We are taking that into consideration, dear. But notice that annular(ring shaped) area above/between the two curved surfaces which is basically the difference of 1/2 circle's area of both radius 6 and 4 units..
Team BV - Kamal Lohia
Team BV - Kamal Lohia
@Koushik98 said:One-fourth portion of each of the two solid spheres having radii 4 units and 6 units are joined together such that the centres of the two spheres coincide as shown in the figure given below. What is the total surface area of the solid thus formed? a)88 pi b)83 pi c)75 pi d)73 pi
a option.
88pi ?
88pi ?