Latest Elitmus Aptitude Question SOLUTION: How many ways the three digits number (XYZ) can be formed in the form XZ?

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How many ways the three digits number (XYZ) can be formed in the form XZ?

Asked In Elitmus Elakkiya Balasubramanian (8 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Suhit  (8 years ago)

Submitted by nishant:- take value of X Y Z in 0 -9
here value of x cant be 0 as its in cent place.

take value of x - 1 to 9
y -- 0 -9
z -- 0 - 9

now put x =1 y = 0 then z = 0 to 9 here we cant use 1 for y and 1 and 0 for z.
1 0 2, 10 3, 10 4, 105, 106, 107, 108, 109
so for x = 9 value possible and for z = 8 ---so total value possible are 72 and in y we can put different 8 values so answer is 9 * 8 * 8 == 64*9= 576

Is this solution Helpfull? Yes (11) | No (6)  

ashish ranjan  (7 years ago)

Total number of 3 digit numbers = 9×10×10 = 900
Total number of numbers in which no digit repeats = 9×9×8 = 648
So the total number of numbers in which at least one digit repeats = 900 - 648 = 252

Is this solution Helpfull? Yes (3) | No   

happy goyal  (7 years ago)

Above question is wrong right question is

A three digit number of the form ‘xyz’ is to be formed such that x > y, x > z, and y < z. How many such numbers are possible if x > 0?

Is this solution Helpfull? Yes (2) | No   

Vikash kumar  (7 years ago)

i think question is wrong
if x

Is this solution Helpfull? Yes (1) | No (1)  

sooraj singh  (7 years ago)

data insufficient

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Abhilasha Srivastava  (7 years ago)

x=1
y=2 z= 3...9 (7 ways)
y=3 z= 4...9 (6 ways)
...
y=8 z= 9 (1 way)

Total : 7+6+...+1

x=2
y=3 z= 4...9 (6 ways)
y=4 z= 5...9 (5 ways)
...
y=8 z= 9 (1 way)

Total : 6+5+...+1

...

x=6
y=7 z= 8,9 (2 ways)
y=8, z=1 (1 way)
Total : 2+1

x=7
y=8 z= 9(1 way)

Total : 1


Required number of ways = ( 7+6+...+1) + (6+5+...+1) + (5+4+3+2+1) + (4+3+2+1) + (3+2+1) + (2+1) + 1
=28+21+15+10+6+3+1
= 84
Qn is x

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prachi  (7 years ago)

I think difference shud be 11 only

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Ramesh Sriram  (7 years ago)

3p2
i.e
only 6 i think

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shreeprakash  (7 years ago)

assuming x cannot be zero
xyz
x,y,z can be 0,1,2,3,4,....9
1)y=0 not possible
1)y=1 x=z=0 not possible
2)y=2 x,z can be filled with{0,1} =1*2= 2case..............{x in 1 way and z in 2{x cannot be zero}}
3)y=3 x,z can be filled with{0,1,2}=2*3=6 case
..
..
9)y=9 x,z can be filled with{0,1,2..8}=8*9=72 cases

Total cases 1*2+2*3+..........n(n+1)=1+n(n+1)(n+2)/3
here n is 8
hence total cases=240 ans

if x can be zero
than no of two digit number with y>z is 9{10,20,30..90}+8{21,31,41..91}+7{32,42....

hence total case=285

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