yes i do have but i would like to give few more time before publishing the answer.. dont want to ruin other's hard work. A strictly increasing no. is one in which each digit exceeds the digit to its left. f.e. 1478 how many strictly inc. numbers are there belowa) one thousandb) one million (XAT kinda question)
A strictly increasing no. is one in which each digit exceeds the digit to its left. f.e. 1478 how many strictly inc. numbers are there belowa) one thousandb) one million (XAT kinda question)
A strictly increasing no. is one in which each digit exceeds the digit to its left. f.e. 1478 how many strictly inc. numbers are there belowa) one thousandb) one million (XAT kinda question)
yes i do have but i would like to give few more time before publishing the answer.. dont want to ruin other's hard work. A strictly increasing no. is one in which each digit exceeds the digit to its left. f.e. 1478 an increasing number is one in which each digit is not exceeded by digit to its left.f.e 3445 how many inc. numbers are there belowa) one thousandb) one million (XAT kinda question)
a) C(12, 9)
b) C(15, 9)
It will be given by no of non-negative integral solutions of the eq
x0 + x1 + x2 + .... + x9 = n,
xr is no of times digit r appears and n is the no of digits in the largest number we can form
Edit:- we need to subtract 1 also, as 000 and 000000 is also counted
yes i do have but i would like to give few more time before publishing the answer.. dont want to ruin other's hard work. A strictly increasing no. is one in which each digit exceeds the digit to its left. f.e. 1478 an increasing number is one in which each digit is not exceeded by digit to its left.f.e 3445 how many inc. numbers are there belowa) one thousandb) one million (XAT kinda question)
@chillfactor's method is faster as it gives the answer with 1 calculation. I used a longer version of partitioning theory only.
Basically the idea is, suppose you want a 4 digit number. No digit can be greater than 9 and none less than 1. So you start from 1 and increment to 9, stopping at 4 points on the way (which are your 4 digits). Hence we need 5 increments totaling up to 8 units (an increment may be zero also since repeated digits are allowed). In other words we want to divide 8 units into 5 spaces or groups such that groups may be empty, which gives (8+4)C4 or 12C4. For 5 digits, 13C5 similarly, and so on...
Here we want up to 6 digits, so 9C1 + 10C2 + 11C3...till 14C6
But 8C0 + (9C1 + 10C2 + 11C3...till 14C6) = 15C6 and 8C0 is 1.
Hence it actually comes to the same 15C6 -1. regards scrabbler
Are you sure its integral solutions and not positive integral solutions ?Let S = {2006, 2007, 2008,…….., 4012}. 'K' denotes the sum of the greatest odd divisor of each of the element of S. Find the value of 'K'.