[arya5139]

Originally Posted by noesis

1. n1, n2, n3 ... n10 are 10 numbers such that n1 > 0 and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number ?

(1) 109
(2) 27
(3) 36
(4) None of these

1> Total number of triplets = (8+7+6+......+1) + (7+6+5+....+1) + (6+5+4+...1) +.......+1
=36+28+21+15+10+6+3+1 = 120 => option 4

is this correct?

[MaskedMenace]

Originally Posted by noesis

1. n1, n2, n3 ... n10 are 10 numbers such that n1 > 0 and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number ?

(1) 109
(2) 27
(3) 36
(4) None of these

2. What is the remainder obtained on dividing

16^101 + 8^101 + 4^101 + 2^101 + 1 by 2^100 +1

1.10c3 = 120

2.

put 2^100 = (-1)

2^4 - 2^3 + 2^2 - 2 + 1

16 - 8 + 4 - 2 + 1

11

[Leonayas]

Sum of all 4 digit numbers without repeatation wil be
(10^3)*(3!)*(1+2+3+4) + (10^2)*(3!)*(1+2+3+4) + (10)*(3!)*(1+2+3+4) + (1)*(3!)*(1+2+3+4)
= (3!)*(1+2+3+4)*(10^3+10^2+10+1)
=6*10*1111

[prakharc]

Originally Posted by chandan.mnnit

can you please explain in a detailed manner!!!

with (n1,n2,x)..8 values of x are possible....n3 to n10
with (n1,n3,x)...7 values of x are possible....and so on

So with 1st number as n1 we have 8+7+6+5+...1 = 36 possible

Similarly with 1st number as 2...
with (n2,n3,x)...7 values of x are possible..and so on
So with 1st numebr as n2 we have 7+6+5+...1 = 28 possible

Similarly 1st number as n3 to n7...

[arya5139]

Originally Posted by chandan.mnnit

A certain calculator has a faulty display which can display only the digits 1,2,3 and 4. find the sum of all the four digit numbers with distinct digits that the calculator can display?

p.s - I have solved this question but by a lengthy method...looking for a smart solution

Total number of 4 digit numbers with distinct digits = 4*3*2*1= 24

=>there will be 6 numbers with unit digit as 1 ,6 numbers with unit digit 2...and so on...
=>sum = 6*( 11110) = 66660

[MaskedMenace]

q. How many 3 digit nos. Are such that it's digits are in increasing order (i.e. From l->r) ?

This was the similar question...posted somedays bak...

{1+2+3+4+5+6+7} + {1+2+3+4+5+6} + {1+2+3+4+5} + { 1+2+3+4} + {1+2+3} + {1+2} + 1

=> 84

or 9c3 = 84

[chandan.mnnit]

Originally Posted by chandan.mnnit

A certain calculator has a faulty display which can display only the digits 1,2,3 and 4. find the sum of all the four digit numbers with distinct digits that the calculator can display?

p.s - I have solved this question but by a lengthy method...looking for a smart solution

Just tried in another way...tell me if its correct approach or not

total number of ways is 4!

in this each digit will occur at each place 4!/4 times that is 6 times

so at units place sum = 6* (1+2+3+4) =60

similarly we can find out for each of the tens, hundreds and thousands place.......Is this approach correct?

[Leonayas]

spot on
sum of numbers formed by 4 digits 1,2,3,4 without repeating is
1000*3!*(1+2+3+4) + 100*3!*(1+2+3+4) + 10*3!*(1+2+3+4) + 1*3!*(1+2+3+4)
= 3! * (1+2+3+4) * 1111
= 6 * 10 * 1111
= 66660

[effort]

Originally Posted by arya5139

1> Total number of triplets = (8+7+6+......+1) + (7+6+5+....+1) + (6+5+4+...1) +.......+1
=36+28+21+15+10+6+3+1 = 120 => option 4

is this correct?

u r absolutely rite

[prateek7563]

Originally Posted by MaskedMenace

q. How many 3 digit nos. Are such that it's digits are in increasing order (i.e. From l->r) ?

This was the similar question...posted somedays bak...

{1+2+3+4+5+6+7} + {1+2+3+4+5+6} + {1+2+3+4+5} + { 1+2+3+4} + {1+2+3} + {1+2} + 1

=> 84

or 9c3 = 84

i didnt understand the solution.