@vijay_chandola said:for unit digit 2, 3, 5 = 5! combinations eachfor 4 and 6 = 2*4! eachtotal = 5!+5!+2*4!+5!+2*4! = 456
oa is 528
@vijay_chandola said:for unit digit 2, 3, 5 = 5! combinations eachfor 4 and 6 = 2*4! eachtotal = 5!+5!+2*4!+5!+2*4! = 456
@raopradeep said:find the number of 6 digit numbers that can be found using 1,2,3,4,5,6 once such that the 6 digit number is divisible by its unit digit (the unit digit is not 1)620456520372
@vijay_chandola said:for unit digit 2, 3, 5 = 5! combinations eachfor 4 and 6 = 2*4! eachtotal = 5!+5!+2*4!+5!+2*4! = 456
@Jackson1 said:I am getting 528...when unit digit is either 2,3,5 or 6 then each number would be divisible by the unit digit..so possible numbers are 5*4*3*2*1*4 = 480when unit digit is 4 then only two digits can be at ten's place i.e 2 and 6.so possible numbers are 4*3*2*1*2*1= 48total numbers are = 480 + 48 = 528..plz tell me where i am getting wrong??
@raopradeep said:@vijay_chandolaoa is 528
@raopradeep said:oa is 528
@raopradeep said:@vijay_chandolaoa is 528
@vijay_chandola said:for unit digit 2, 3, 5 = 5! combinations eachfor 4 and 6 = 2*4! eachtotal = 5!+5!+2*4!+5!+2*4! = 456
there are N men sitting around acircular table at N distinct points . every possible pair of men except the ones sitting adjacent to each other sings a 2 minute song one pair after other. of total time taken is 88 min. find N ?
@scrabbler said:Which is what I was getting...par woh options mein tha hi nahin regardsscrabbler
@raopradeep said:there are N men sitting around acircular table at N distinct points . every possible pair of men except the ones sitting adjacent to each other sings a 2 minute song one pair after other. of total time taken is 88 min. find N ?810911
@abhishek.2011 said:bro 6 k case main_ _ _ _ _ 6as sum of numbers is 21 so any combination will be divisible by 3as last digit is even so divisible by 2hence by 6so why 2*4! ???????
@vijay_chandola said:Kadbury offers a packet of 5 chocolates at the list price of 4 chocolates and on purchasing 19 such packets gives one packet absolutely free. A trader receives 20 packets of the chocolates in the offer and sells each chocolate at its list price. What is his net percentage profit?
(a) 24% (b) 31.58% (c) 35% (d) 53.75%
@raopradeep said:there are N men sitting around acircular table at N distinct points . every possible pair of men except the ones sitting adjacent to each other sings a 2 minute song one pair after other. of total time taken is 88 min. find N ?810911
ABC is the internal bisector of the angle meets BC at D. If Ab 2root3 AC is 4root3, angle A is 60 degrees, find AD.
@scrabbler said:11 hoga....N(N-3)/2 = 44This is a CAT actual question I guess? From early 2000s...regardsscrabbler
@veertamizhan said:ABC is the internal bisector of the angle meets BC at D. If Ab 2root3 AC is 4root3, angle A is 60 degrees, find AD.
@veertamizhan said:ABC is the internal bisector of the angle meets BC at D. If Ab 2root3 AC is 4root3, angle A is 60 degrees, find AD.
Bahut time laga diya :(zada has to distribute 15 choclates among her 5 kids(a,b,c,d,e) . such that "a" gets atleast 3 and at most 5 chocolates .in how many ways she can do this ?
@raopradeep said:bhai this eqtn means i think n * (n-3) means the man himself and his two neighbors including him =88 min song rectify if iam wrong