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5 Jul 2007 00:29:40 IST
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You made a slight mistake. You havent considered the cases where a single variable can take all the four numbers, ie, (7x5x3x2), 1, 1. etc.
So the solution is :
Case one : We select two numbers out of the given 4 in 4c2 ways, and we can assign the new product and the remaining 2 numbers to the 3 variables in 36 ways, as cooldude rightly pointed out.
Case two: We select three out of the given four numbers, in 4c3 = 4 ways, and let the product be p. Then, [a, p, 1] (a is the remaining number) can be arranged in 6 ways, ie total no of ways here is 6x4 = 24 ways.
Case three: We select all four of the numbers in 1 way, and assign it to one variable, ie, it becomes [p, 1, 1]. This arrange ment can be done in 3!/2! ways = 6/2 = 3 ways.
Total number of solutions is 36+24+3 = 63 integral solutions.
The general algorithm for such problems is:
1. Let the problem be something like abcde....z = n (n
I)
2. Prime factorize n.
3. Now, simply find the number of ways in which you can arrange the individual factors of n among the given variables.
4. Remember to include all cases. For example, if you have something like abc = 2x3x7, then you will have 3 cases. The first case will be when a variable, ie, a, b, or c takes a single factor, ie, it can take 2 or 3 or 7. Here there are 3! ways for that to happen.
The second case will be when a variable takes the product of 2 factors, ie, here, it will be like (2x3), 7, 1. So the number of arrange ments here will be 3! again.
The third case will be when a variable takes the product of 3 factors, ie (2x3x7), 1, 1. Here no of arrangements is 3!/2! (Just like arranging letters of a word where the words are repeated).
Keep doing this procedure till you exhaust all cases.
Hope it helped!
So the solution is :
Case one : We select two numbers out of the given 4 in 4c2 ways, and we can assign the new product and the remaining 2 numbers to the 3 variables in 36 ways, as cooldude rightly pointed out.
Case two: We select three out of the given four numbers, in 4c3 = 4 ways, and let the product be p. Then, [a, p, 1] (a is the remaining number) can be arranged in 6 ways, ie total no of ways here is 6x4 = 24 ways.
Case three: We select all four of the numbers in 1 way, and assign it to one variable, ie, it becomes [p, 1, 1]. This arrange ment can be done in 3!/2! ways = 6/2 = 3 ways.
Total number of solutions is 36+24+3 = 63 integral solutions.
The general algorithm for such problems is:
1. Let the problem be something like abcde....z = n (n
I) 2. Prime factorize n.
3. Now, simply find the number of ways in which you can arrange the individual factors of n among the given variables.
4. Remember to include all cases. For example, if you have something like abc = 2x3x7, then you will have 3 cases. The first case will be when a variable, ie, a, b, or c takes a single factor, ie, it can take 2 or 3 or 7. Here there are 3! ways for that to happen.
The second case will be when a variable takes the product of 2 factors, ie, here, it will be like (2x3), 7, 1. So the number of arrange ments here will be 3! again.
The third case will be when a variable takes the product of 3 factors, ie (2x3x7), 1, 1. Here no of arrangements is 3!/2! (Just like arranging letters of a word where the words are repeated).
Keep doing this procedure till you exhaust all cases.
Hope it helped!
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select 2 of these no.s(can be done in 4c2=6ways) and multiply them.let their product be p.then {a b c}={other 2 numbers,p}.each of a,b and c can be alloted to all 3 numbers.This again can be done in 6 ways.
hence total 6X6=36 solutions.