@abhi_14 said:In how many ways can 10 identical presents b distributed among 6 children so tht each gets atleast one present15C516C69C56^10NOTPlease share the detail approach
a + b + c + d + e + f = 10
(n - 1)c(k - 1)----> (10 - 1)c(6 - 1) = 9c5 ?
Q 21, 22 :-)
The cost of digging a pit was 1347. how much will be the cost if length of the working day increased by 1/20 of the former period?
@vbhvgupta said:The cost of digging a pit was 1347. how much will be the cost if length of the working day increased by 1/20 of the former period?
Cost = k*t
t' = 21l/20
C' = 21/20 *C = 1414.35 ? ..
@abhi_14 said:@scrabbler could u pls share the approach to tackle all the four cases i.e.:
1. identical balls, identical basket
This has to be listed out....slightly time-consuming it is
@abhi_14 said:
2.identical balls, different basket
3.different basket, identical ball
Identical cases, different order 😉 ? I assume one should be identical basket different ball - that would be the most difficult of them all - first would have to list as in case 1, then arrange for each case separately and add.
Different basket identical ball is when we do theory of partitioning. Long story hai, but in brief, to divide among n (distinct) baskets we need to put n (identical) partitions among the balls, so effectively we are arranging n + (r-1) objects out of which n are identical and the remaining (r-1) are also identical giving us (n+r-1)!/ n! (r-1)! which is (n+r-1)C(r-1). This can be easily modified to account for conditions like "no group is empty"
@abhi_14 said:
4.different basket, different ball
Here we use the n^r logic...say 5 balls to 8 baskets will be 8^5 ways
@abhi_14 said:or if u knw any good source fr this funda, could u pls share this, I hv lot f confusion related to this topicPS: Ur blog seems to b really good, great initiative
Thanks 😃
In fact I will be tackling all the above cases in more detail there at some future point (around a month from now at least, since I need to cover the easier "distinct" cases first...if you can have patience till then 😁 Nahin to there are several textbooks I guess which would deal with this....don't know any names, but surely NCERT mein hoga...)
regards
scrabbler
@vbhvgupta said:The cost of digging a pit was 1347. how much will be the cost if length of the working day increased by 1/20 of the former period?
1414.35 ?
let initial days be x
so for x days ----> 1347
for 21/20*x days ----> 1347*21/20 = 1414.35
@vbhvgupta said:The cost of digging a pit was 1347. how much will be the cost if length of the working day increased by 1/20 of the former period?
1414.3 ?
@ScareCrow28 said:Cost = k*tt' = 21l/20 C' = 21/20 *C = 1414.35 ? ..
no change IN cost? even though no explicit relation is given bw cost and time..even though work is same for both cases..and payment should be according to work..
only a guess
@Ani1308 said:no change IN cost? even though nomexplicitnrelation is given bw cost and time..only a guess
There is, in fact, no relation given between cost and time, so we can only assume or guess! 😃 I guessed it ( cost) to be directly proportional to time.
@scrabbler said:A and B? Trying orally again
:o orally? Hows that..? Especially the 22nd one..?
Please teach me the tric bro 
. Am wasting natural resources ..
. Am wasting natural resources ..@albiesriram said:orally? Hows that..? Especially the 22nd one..? Please teach me the tric bro . Am wasting natural resources ..
22nd:
First figure I imagined a vertical line joining the midpoints of the two horizontal sides. So it forms 8 equal area triangles. so the shaded area is 1/4th of the square.
Second figure I drew a diagonal from bottom left to top right. So the square is cut in two triangle, in each of which two medians are drawn and hence in each case the shaded area is 1/3 of the area of that triangle. So overall shaded area is 1/3rd that of square. So difference is 1/3 - 1/4 = 1/12 of area of square = 1/12 * 36 = 3.
Is it correct?
21st is easier, usme 4 * 6 = 24 to aana hai, do option bachte hai, ab usmein check maarna hai shaded area kitna hai...usmein calculation error ho sakta hai aur kuch nahin.
regards
scrabbler
First figure I imagined a vertical line joining the midpoints of the two horizontal sides. So it forms 8 equal area triangles. so the shaded area is 1/4th of the square.
Second figure I drew a diagonal from bottom left to top right. So the square is cut in two triangle, in each of which two medians are drawn and hence in each case the shaded area is 1/3 of the area of that triangle. So overall shaded area is 1/3rd that of square. So difference is 1/3 - 1/4 = 1/12 of area of square = 1/12 * 36 = 3.
Is it correct?
21st is easier, usme 4 * 6 = 24 to aana hai, do option bachte hai, ab usmein check maarna hai shaded area kitna hai...usmein calculation error ho sakta hai aur kuch nahin.
regards
scrabbler
@chillfactor cant be zero dude....check ur method again....remainder would undoubtely be zero bt nt the quotient. i ll post the answer as soon as i get it.. thanks anyways.
@shattereddream said:The half-life of a substance is 1 hr, i.e. at the end of 1 hr, its quantity reduces to one-half of what it was at the beginning of the hour. In how many hours will the quantity become less than 1% of the initial qua
OA is 7
@vbhvgupta said:The cost of digging a pit was 1347. how much will be the cost if length of the working day increased by 1/20 of the former period?
Total cost = 1347 x(9/8 x (20/21)
= 1443.214
A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.)
a.4
b.28
c.10
d.16
e.22
a.4
b.28
c.10
d.16
e.22
@shattereddream said:A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
22 ?
@shattereddream said:A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till €˜n €™ terms. What is the value of (T5040 + T5042 + T5044 + T5046)? (Tn is the nth term of the given sequence.) a.4 b.28 c.10 d.16 e.22
T2, T6, T12, T20, T30 are all sevens, so in this pattern.
'n' th term of the above series need to be found
S = 2 + 6 + 12 + 20 + 30 + ....
S = 0 + 2 + 6 + 12 +....
T(n) = 2 + 4 + 6 + 8 + 10 ..... = 2*(n*(n+1)/2) = k*(k+1)
T( k * (k+1) ) = 7
k = 70 => T(4970) = 7
Now T( k *(k +1) ) is followed by (k + 1) ones.
=> T(4971) to T(4970 + 71) = 1
So, T(5040) = 1, T(5042), T(5044), T(5046) = 7
=> Sum = 1 + 7 +7 + 7 = 22 ? :(