there are 10 stalls in a ship and there are cow's calves and horses to be transported. in how many ways can the ship load be made if cow and horses are not less than 10 but maximum no. of calves is 7.
Akshita (student) (108 Points)
04 June 2012there are 10 stalls in a ship and there are cow's calves and horses to be transported. in how many ways can the ship load be made if cow and horses are not less than 10 but maximum no. of calves is 7.
intermediate(ipc)course
(no)
(1460 Points)
Replied 04 June 2012
pl give answer. because in cpt exam is necessary.
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 04 June 2012
If the cows and horses are not less than 10 then it implies that there are exactly 10 total animals( otherwise we will be forced to keep two animals in one stall!!!!!) so the number calves can vary from 0 to 7 and no of horses can vary from 10 to 3 respectively. the problem does not say whether each animal is unique or all horses are the same and so are all the calves I will try to solve both the cases
Case 1: each animal is unique(1 horse is different from another horse and 1 calf is differernt from another)
now there are 8 (0+10 ; 1+9 ; 2+8 ; 3+7 ; 4+6 ; 5+ 5; 6+4 ; 7+3)varying number of compositions of calves and horses which can be treated as 10 different objects to be arranged in 10 places so answer is 8*10! = 29030400
Case 2: each animal is not distinct from any animal of same species.
10c10 + 10c9*1c1+10c8*2c2+10c7*3c3 + ....+10c3*7c7 = 968 as the answer
the given answer would only begin to make sense in a fallacious way if certain assumptions are made. most important being the total number of animals can be anything between 0 and 10. now someone can say that there can be 3 states possible for 7 stalls in which calves can be placed (i.e empty stall, stall with horse, stall with calf) and the other 3 stalls can have 2 states (empty stall and stall with horse). so total number of possibilities is 2^3*3^7 (I am only speculating here as to what they might think be the method in which we should solve the problem). the above solution also assumes that the horses and calves are distinct from each other. The reasoning sounds very plausible but is very wrong. If we remove the restriction saying the number of animals is 10 and allow it to vary between 0 and 10 the problem will become very messy. I solved the variation in which the restriction of 10 animals is removed for both the cases and the answer comes to this
case 1: all animals are unique: 78682611 (4497 times the given answer)
case 2: animals of same species are indistinguishable : 58848 ( 3 times the given answer)
I am reasonably confident that my answers are correct as such if you want the solutions for the above variation just ask me
Akshita
(student)
(108 Points)
Replied 05 June 2012
well i thought there is some mistake in the question.. for in the starting it is written cow's calves and in the next line.. they are taking it as three.. cows, calves and horses.. what would be the answer if it is just cow's calves and horses?
Tejaswi Kasturi
(student-cpt)
(427 Points)
Replied 05 June 2012
the no. of animals is not less than 10. so we should assume that no. of animals is 10 as we can't put more than 1 animal in a stall and we have 10 stalls. now we don't know the composition of the group of animals. i.e. we don't know whether there are no calves or if there is 1 calf or there are 2 or 3 or 4 or .. calves. we don't know how many calves are there or how many horses are there. let x be the number of calves and y be the number of horses. we only know that
x+y= 10 and x<=7 and that y can be anything between 0 and 10 so we try write all the possible combinations.
combination 1(c1): 0 calves and 10 horses
combination 2(c2): 1 calf and 9 horses
combination 3(c3): 2 calves and 8 horses
combination 4(c4): 3 calves and 7 horses
combination 5(c5): 4 calves and 6 horses
combination 6(c6): 5 calves and 5 horses
combination 7(c7): 6 calves and 4 horses
combination 8(c8): 7 calves and 3 horses
no more combinations are possible since the no: of calves cannot be greater than 7. let us denote combination n by cn.
now they also did not say whether to assume that the one horse is different from another or not, i.e. if a and b are two horses whether ab and ba are same arrangements or not. they would be same if you cannot distinguish one horse from another. It would not be so if you can distinguish one horse from another. similarly they did not mention whether one calf is different from another calf. we will try to solve both the cases i.e. case 1: no two animals are same case 2: horses are indistinguishable among themselves and calfs are indistinguishable among themselves.
CASE 1: for c1 combination we can arrange 10 horses in 10 places in 10! ways. for c2 combination we can arrange 9 horses in 10 places in 10P9 ways and 1 calf in the remaining place i.e 10p9*1 = 10! ways. for c3 combination we can arrange 8 horses in 10P8 ways i.e. 10!/2! ways and the remaing 2 calves in 2 places in 2! ways so total no. of ways is (10!/2!)*2! = 10! ways. similarly for the rest of combinations. summing them all we get the number of arrangements as 8*10! = 29030400
CASE 2:
for c1 combination there is only 1 way in which we arrange the horses as they are all same. 1
for c2 combination we can select 1 place for the calf in 10 ways and arrange the horses in remaining places 10
for c3 combination we can select 2 places for the calves in 10c2 ways and arrange the horses in remaining places 45
for c4 combination we can select 3 places for the calves in 10c3 ways and arrange the horses in remaining places 120
for c5 combination we can select 4 places for the calves in 10c4 ways and arrange the horses in remaining places 210
for c6 combination we can select 5 places for the calves in 10c5 ways and arrange the horses in remaining places 252
for c7 combination we can select 6 places for the calves in 10c6 ways and arrange the horses in remaining places 210
for c8 combination we can select 7 places for the claves in 10c7 ways and arrange the horses in remaining places 120
so total is 1 + 10 +45 + 120 +210 +252 +210 +120 = 968. so total no. of ways is 968.
sruthy
(student)
(70 Points)
Replied 07 June 2012
@ tejaswi kasturi :
you will surely pass ur cpt xam dis tym.....
all d best ......all z welll...... :) :D
wishes from,
sruthybaskaran. :) :D
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