In how many ways can 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates?
You can solve this question by using the reverse method for distribution- that is picking. First distribute 6 chocolates each to all the three students. Now they have 18 chocolates in total. Now you need to pick 6 chocolates out of these 3 groups which is the solution of a + b + c = 6 = 8!/6!2! = 28
Find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12.
As the first digit cannot be 0, we require the coefficient of x12 in the product (x + x2 + x3 + ... + x9)(1 + x + x2 + x3 + ... + x9)3 = coefficient of x11 in the product = (1 - x9)/(1 - x) × (1 - x10)3/(1 - x)3 = (1 - x9)(1 - x30 - 3x10 + 3x20)(1 - x)-4 = 332 (expanding (1 - x)-4). I hope the answer is correct.
Find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12.
wy do we take the co-efficient of x^11 as the answer. i do not understand wy are u multiplying those two expressions in the first place . please explain the logic behind it.
but i did try another way to find the answer and it seems correct to me. please verify.
the sum of the numbers should be 12 and i consider them as 12 balls. and since it is a 4 digit number i consider it as four different baskets(a,b,c,d). Now, the first digit cannot be zero so i put one ball in the first basket(a){let me call the rest of the balls that go to basket one as a'}. therefore the problem reduces to the number of ways of dividing these 11 balls into the four baskets with three important caveats:
1. all the cases where all the 11 balls have been put in the same basket have to be excluded
2. all the cases where 10 balls have been put in the same basket have to be excluded
3. all the cases where 9 balls have been put in the first basket has to be excluded(since it ll make the value as 10 for the first basket a, this is invalid since the maximum that we can put in a basket is 9).
Now for the calculations,
no. of ways of dividing 11 balls among 4 baskets is:
a'+b+c+d = 11, (n+r-1)C (r-1)= (11+4-1)C (4-1) = 14 C 3 = 364
Now for case 1: if all eleven are put into one basket it is wrong since the maximum that can be there in a basket is 9, therefore we exclude this. this is nothing but arranging 11,0,0,0 which is equal to
4!/3! = 4 ways
similarly for case 2: this is nothing but arranging 10,1,0,0 which is equal to
4!/2! = 6 ways
for case 3: we need to consider only those cases where we have 9 in the first basket, and they are the number of ways of arranging
9,1,1,0 (with nine fixed in the first place) = 3!/2! = 3 ways
9,2,0,0 (with nine fixed in the first place) = 3!/2! = 3 ways
hence the total is 364 - (4+6+3+3) = 364-16 = 348
You can solve this question by using the reverse method for distribution- that is picking. First distribute 6 chocolates each to all the three students. Now they have 18 chocolates in total. Now you need to pick 6 chocolates out of these 3 groups which is the solution of a + b + c = 6 = 8!/6!2! = 28
Find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12.
As the first digit cannot be 0, we require the coefficient of x12 in the product (x + x2 + x3 + ... + x9)(1 + x + x2 + x3 + ... + x9)3 = coefficient of x11 in the product = (1 - x9)/(1 - x) × (1 - x10)3/(1 - x)3 = (1 - x9)(1 - x30 - 3x10 + 3x20)(1 - x)-4 = 332 (expanding (1 - x)-4). I hope the answer is correct.
Find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12.
wy do we take the co-efficient of x^11 as the answer. i do not understand wy are u multiplying those two expressions in the first place . please explain the logic behind it.
but i did try another way to find the answer and it seems correct to me. please verify.
the sum of the numbers should be 12 and i consider them as 12 balls. and since it is a 4 digit number i consider it as four different baskets(a,b,c,d). Now, the first digit cannot be zero so i put one ball in the first basket(a){let me call the rest of the balls that go to basket one as a'}. therefore the problem reduces to the number of ways of dividing these 11 balls into the four baskets with three important caveats:
1. all the cases where all the 11 balls have been put in the same basket have to be excluded
2. all the cases where 10 balls have been put in the same basket have to be excluded
3. all the cases where 9 balls have been put in the first basket has to be excluded(since it ll make the value as 10 for the first basket a, this is invalid since the maximum that we can put in a basket is 9).
Now for the calculations,
no. of ways of dividing 11 balls among 4 baskets is:
a'+b+c+d = 11, (n+r-1)C (r-1)= (11+4-1)C (4-1) = 14 C 3 = 364
Now for case 1: if all eleven are put into one basket it is wrong since the maximum that can be there in a basket is 9, therefore we exclude this. this is nothing but arranging 11,0,0,0 which is equal to
4!/3! = 4 ways
similarly for case 2: this is nothing but arranging 10,1,0,0 which is equal to
4!/2! = 6 ways
for case 3: we need to consider only those cases where we have 9 in the first basket, and they are the number of ways of arranging
9,1,1,0 (with nine fixed in the first place) = 3!/2! = 3 ways
9,2,0,0 (with nine fixed in the first place) = 3!/2! = 3 ways
hence the total is 364 - (4+6+3+3) = 364-16 = 348
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