let me explain 1 case...x=0 wala..here y+z = 30subject to constrain -> y,z>3y+z -> (4+26),(5+25),...,(26+4)so total = (26-4)+1 = 23...but minus where both are equal, hence 22 solutions...abb batao kuch samajh nahi aya toh??
got it bhai aaj dimag pura kharab ho gaya hai thanks for having patience :)
This is the first time I have seen you banging your head!
I listen to soft old music rather than rock...so headbanging is not my thing ;) Been a long day today...and I started off with a migraine 😛 So not in the most patient of moods I will admit 😞 no offense meant...
All possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?
All possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?OPTIONS1) 632100 2) 622110 3) 633110 4) 642100 QOD TF...
500th or 5000th? There are over 2000 starting with 9 only I am getting 😞 Ya phir kuch galat kar raha hoon... regards scrabbler
All possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?OPTIONS1) 632100 2) 622110 3) 633110 4) 642100 QOD TF...
633110. Bahut ganda method, please don't ask me to type it out 😞 Trying to think of a better approach...but not likely, too sleepy... Edit: Typed it after all. At least part of it. :P
All possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?OPTIONS1) 632100 2) 622110 3) 633110 4) 642100 QOD TF...
i am getting 633110lets start with 1 as lsb100000 till 111111 -> 6 such numbers...200000 till 222222 -> 21 such numbers..300000 till 333333 -> 56 numbers..400000 till 444444 -> 126 numbers...500000 till 555555 -> 252 numbers...up till here we have got 462 numbers..600000 -> 1 number61xxxx -> 5 numbers62xxxx -> 15 numbers..63xxxx -> 35 numbers..this amounts to 482 numbers...abb aese hi kar kar ke... 633110 pe aya...pata nahi kya kara hai....ajeeb question.. @scrabbler pls check karo...
Similar approach, wohi answer, took too much time though :'( Koi accha method anyone? regards scrabbler
Similar approach, wohi answer, took too much time though :'( Koi accha method anyone?regardsscrabbler
mereko ek bada interesting sa pattern mila hai iska....i know there has to be a logical explanation behind it...which i am somehow not able to pen down coz mereko samjh nahi aa raha kaise bataun...but
1xxxxx -> now the 5 x's can be filled with either 1 or 0 so total possible such numbers = (number of x's + total possible digits that can fill those x's - 1)C(total possible digits that can fill those x's - 1) [here C is combination wala nCr wala]...
so 1xxxxx -> (5+2-1)C(2-1) = 6c1 = 6 numbers
2xxxxx -> (5+3-1)c(3-1) = 7c2 = 21 numbers...
abb isko words mein samjhane ki koshish karunga, agar kar paya toh...
actually it is an extension of a+b+c+d+e+f+g+h+i = k
where a is the number of time 1 occurs etc etc....
mereko ek bada interesting sa pattern mila hai iska....i know there has to be a logical explanation behind it...which i am somehow not able to pen down coz mereko samjh nahi aa raha kaise bataun...but1xxxxx -> now the 5 x's can be filled with either 1 or 0 so total possible such numbers = (number of x's + total possible digits that can fill those x's - 1)C(total possible digits that can fill those x's - 1) [here C is combination wala nCr wala]...so 1xxxxx -> (5+2-1)C(2-1) = 6c1 = 6 numbers2xxxxx -> (5+3-1)c(3-1) = 7c2 = 21 numbers...abb isko words mein samjhane ki koshish karunga, agar kar paya toh...actually it is an extension of a+b+c+d+e+f+g+h+i = kwhere a is the number of time 1 occurs etc etc....
mereko ek bada interesting sa pattern mila hai iska....i know there has to be a logical explanation behind it...which i am somehow not able to pen down coz mereko samjh nahi aa raha kaise bataun...but1xxxxx -> now the 5 x's can be filled with either 1 or 0 so total possible such numbers = (number of x's + total possible digits that can fill those x's - 1)C(total possible digits that can fill those x's - 1) [here C is combination wala nCr wala]...so 1xxxxx -> (5+2-1)C(2-1) = 6c1 = 6 numbers2xxxxx -> (5+3-1)c(3-1) = 7c2 = 21 numbers...abb isko words mein samjhane ki koshish karunga, agar kar paya toh...actually it is an extension of a+b+c+d+e+f+g+h+i = kwhere a is the number of time 1 occurs etc etc....
Yeah it uses nCr values...
See suppose I want all the 6-digit numbers in non-increasing order. What I can do is, since no digit is more than 9, start with =9 and write 7 variables surrounding the digits as follows: a_b_c_d_e_f_g
Where a, b, c, d, e, f are the reductions from 1 digit to the next starting from 9 and ending in 0.
So, for example, for the number 8 7 7 4 3 1 the values of a, b, c, d, e, f and g will be 1103121, as 9 - 1 = 8 - 1 = 7 - 0 = 7 - 3 = 4 - 1 = 3 - 2 = 1 - 1 = 0. Similarly for 643100 it will be 3212100 as 9 - 3 = 6 - 2 = 4 - 1 = 3 - 2 = 1 - 1 = 0 - 0 = 0 - 0 = 0.
Note that the sum of abcdefg will always be 9! So I can solve the number of ways by putting 9 digits into 7 spaces i.e.9+6C6 = 15C6 ways.
If I want numbers starting with 9, I have 6 variables only and I will get 14C5 ways = 2002 (which I mnentioned in previous post)
If I want starting with 5, 4, 3, 2 and 1, they will be 10C5 + 10C5 + 9C5 + 8C5 + 7C5 + 6C5 = 461, then continuing similarly can go till 500. But by God time lagta hai 😞
Also, if you're still reading till here, I am officially impressed.Thank you.
Yeah it uses nCr values...See suppose I want all the 6-digit numbers in non-increasing order. What I can do is, since no digit is more than 9, start with =9 and write 7 variables surrounding the digits as follows:a_b_c_d_e_f_gWhere a, b, c, d, e, f are the reductions from 1 digit to the next starting from 9 and ending in 0.So, for example, for the number 8 7 7 4 3 1 the values of a, b, c, d, e, f and g will be 1103121, as 9 - 1 = 8 - 1 = 7 - 0 = 7 - 3 = 4 - 1 = 3 - 2 = 1 - 1 = 0. Similarly for 643100 it will be 3212100 as 9 - 3 = 6 - 2 = 4 - 1 = 3 - 2 = 1 - 1 = 0 - 0 = 0 - 0 = 0.Note that the sum of abcdefg will always be 9! So I can solve the number of ways by putting 9 digits into 7 spaces i.e.9+6C6 = 15C6 ways.If I want numbers starting with 9, I have 6 variables only and I will get 14C5 ways = 2002 (which I mnentioned in previous post)If I want starting with 5, 4, 3, 2 and 1, they will be 10C5 + 10C5 + 9C5 + 8C5 + 7C5 + 6C5 = 461, then continuing similarly can go till 500. But by God time lagta hai Also, if you're still reading till here, I am officially impressed.Thank you.regardsscrabbler
15C6 ways.If I want numbers starting with 9, I have 6 variables only and I will get 14C5 ways = 2002 (which I mnentioned in previous post)If I want starting with 5, 4, 3, 2 and 1, they will be 10C5 + 10C5 + 9C5 + 8C5 + 7C5 + 6C5 = 461, then continuing similarly can go till 500. But by God time lagta hai Also, if you're still reading till here, I am officially impressed.Thank you.regardsscrabbler
mereko toh jitna meine socha usme hi itna time lagg gaya....ye question exam time chhodna jayda prefer karta....until and unless someone comes up with a shortcut, and i happen to remember that shortcut there...
It's actually pretty quick once you internalise it. At least, the "find all possible 5 digit numbers where digits are non-increasing" kind of questions....this was another level of brutality :P regards scrabbler
Yeah it uses nCr values...See suppose I want all the 6-digit numbers in non-increasing order. What I can do is, since no digit is more than 9, start with =9 and write 7 variables surrounding the digits as follows:a_b_c_d_e_f_gWhere a, b, c, d, e, f are the reductions from 1 digit to the next starting from 9 and ending in 0.So, for example, for the number 8 7 7 4 3 1 the values of a, b, c, d, e, f and g will be 1103121, as 9 - 1 = 8 - 1 = 7 - 0 = 7 - 3 = 4 - 1 = 3 - 2 = 1 - 1 = 0. Similarly for 643100 it will be 3212100 as 9 - 3 = 6 - 2 = 4 - 1 = 3 - 2 = 1 - 1 = 0 - 0 = 0 - 0 = 0.Note that the sum of abcdefg will always be 9! So I can solve the number of ways by putting 9 digits into 7 spaces i.e.9+6C6 = 15C6 ways.If I want numbers starting with 9, I have 6 variables only and I will get 14C5 ways = 2002 (which I mnentioned in previous post)If I want starting with 5, 4, 3, 2 and 1, they will be 10C5 + 10C5 + 9C5 + 8C5 + 7C5 + 6C5 = 461, then continuing similarly can go till 500. But by God time lagta hai Also, if you're still reading till here, I am officially impressed.Thank you.regardsscrabbler
This is the first time I have seen you banging your head! 
