30First thing is to note that the upper limit on the common difference of the sequences that can be formed of length 10 is 11.
Now,
for any sequence with a common difference d and first term 1.
a10=1+9d
Now if the first term of the sequence is increased by 1, the 10th term also increases by 1.
So, we can go on increasing the first term unless the last term becomes 100.
Now, there'll be quite a few cases,
Common difference zero - Clearly, 100 A.P.s are formed.
Common difference one - The last sequence will start from 91. Therefore, 91 A.P.s can be formed.
Common difference two - We consider the first sequence. The 10th term will 19. If the last term increases by 81, it reaches 100, and the first term reaches 82. Hence 82 A.P.s can be formed.
and so on ..
Another thing that can be noted is that as the common difference increases by 1, the no. of sequences decrease by 9 as there is a restriction on the first term.
So the answer should be = 100+91+82+73+64+55+46+37+28+19+10+1 = 606
1ur answer is way too less than the given answer
341perhaps because he was considering only APs
1
chinmaya
·2011-06-03 03:34:05
btw sir it is a doubt[1]..if u can plz help
341
Hari Shankar
·2011-06-03 03:48:51
Hint
For any choice of ten elements from the set, there is only one way of arranging them to obtain a non-decreasing sequence.
Let xi represent the number of times the integer i appears in the given sequence.
Then xi+x2+...+x100 = 10
1
chinmaya
·2011-06-03 04:47:47
thank you got it now!
so answer is coming 109C10
30
Ashish Kothari
·2011-06-03 08:33:21
I completely misunderstood the question! sigh! [2]
62That is ok.. but now solve thsi one.. it is simpler
1
chinmaya
·2011-06-03 08:46:30
sir solve ho to gaya hai na??...i am getting 109C10 as the answer
prophet sir's hint itself was the complete solution[3]
30
Ashish Kothari
·2011-06-03 09:14:42
"10?" sir isn't the answer 109C10?
71Chinamaya And Ashish, explain this expression :
"Then xi+x2+...+x100 = 10 " , Why did Prophet sir mention this expression only and why not he said that it is 100C10 ?
30
Ashish Kothari
·2011-06-04 10:45:00
Because it isn't 100C10. See we need to form an arbitrary sequence from the set of given integers and there is no restriction on the selection .. (thats exactly the mistake that I did at first) So, if we consider the number of times each integer comes and denote it by xi, i being the integer, then surely the sum of all such xis will be 10. Now, since xis≥0, the no. of integral values of xis in their sum evaluates to 109C99. [1]
1@vivek take it like this,
the number of ways arranging x1 + x2 +x3....+x100 = 10 is 109c9.
1the answer is the number of ways of selecting any 10 random numbers from 1 to 100 (any number may be repeated any number of times)
as for any given such selection we can construct a unique sequence like the one demanded.
so your solution is the coefficient of x10
in the expansion of (1+x+x2+x3+x4...)*(1+x+x2+x3+x4...)*(1+x+x2+x3+x4...)..100 times
ith bracket gives a power of x which indicates the occurrence of 'i' in the selection.
coeff. of x10 in (1+x+x2+x3+x4...)100=1/(1-x)100
cheers!
