Binomials

 Posts: 1
 Joined: Sat Oct 10, 2015 2:03 am
Re: Binomials
If you understand case 1, then you should understand case 2.
If there is else any trouble: TRY thinking what will happen if the highest power of 3 with magnitude 3?
(Let me know if you don't understand case 1)
If there is else any trouble: TRY thinking what will happen if the highest power of 3 with magnitude 3?
(Let me know if you don't understand case 1)
 nahin munkar
 Posts: 81
 Joined: Mon Aug 17, 2015 6:51 pm
 Location: banasree,dhaka
Re: Binomials
[I assume,u understand case1,as u wanted case2's explaination.]..
Case2 mainly tells that,u've to take xyz such that the product is not a multiple of 9, but a multiple of 3(as case2,we find out when the product is neither multiple of 3 nor 9 in case1).So, we r now finding out the quantity of product of x,y,z is a multiple of 3,not 9.So,we can't take 9 (from 110)as if we take it, xyz will be a multiple of 9. Now, we have 3 choices.if we choice both 3(3^1) & 6(3*2),then again, 3*6=18 will be multiple of 9.So, we can't take (9) & (both 3 , 6).for this,we have to take one of 3 & 6 in case 2.(so 1st choice taken & we can take it c(2,1)ways )And we have 2 choices remaining.we have to select the 2 choices from other 7 numbers(1,2,4,5,7,8,10)[& we can take it c(7,2)ways]. So, we get total ways  c(2,1)*c(7,2)=2*21=42.By this, we get total 42 xyz whose product is multiple of 3, not higher power than 1 of base 3(like 9,27,..etc..).
And, then after case2, 35(from case1) and 42(from case2) are added and then substracted from 120[c(10,3)total] .At last,probability comes (1203542)/120 or,43/120 as stated.
Case2 mainly tells that,u've to take xyz such that the product is not a multiple of 9, but a multiple of 3(as case2,we find out when the product is neither multiple of 3 nor 9 in case1).So, we r now finding out the quantity of product of x,y,z is a multiple of 3,not 9.So,we can't take 9 (from 110)as if we take it, xyz will be a multiple of 9. Now, we have 3 choices.if we choice both 3(3^1) & 6(3*2),then again, 3*6=18 will be multiple of 9.So, we can't take (9) & (both 3 , 6).for this,we have to take one of 3 & 6 in case 2.(so 1st choice taken & we can take it c(2,1)ways )And we have 2 choices remaining.we have to select the 2 choices from other 7 numbers(1,2,4,5,7,8,10)[& we can take it c(7,2)ways]. So, we get total ways  c(2,1)*c(7,2)=2*21=42.By this, we get total 42 xyz whose product is multiple of 3, not higher power than 1 of base 3(like 9,27,..etc..).
And, then after case2, 35(from case1) and 42(from case2) are added and then substracted from 120[c(10,3)total] .At last,probability comes (1203542)/120 or,43/120 as stated.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss
Re: Binomials
Thanks for taking the time to explain Nahin, I learned a lot from your explanation.