Math Problems
#Problems #Quest37
#Solutions | #Quest37
(1+x)ⁿ=C(n;0)*1+C(n;1)*1*x+C(n;2)*1*x²+...+C(n;n-1)*x^(n-1) +C(n;n)*x^n
n(1+x)^(n-1)=C(n;1)+2C(n;2)*x+3*C(n;3)x²+...+(n-1)*C(n;n-1)*x^(n-2)+n*C(n;n)*x^(n-1)
x=1
C(n;1)+2*C(n;2)+3*C(n;3)+...+n*C(n;n)=n*2^(n-1)
Author:Anar
(1+x)ⁿ=C(n;0)*1+C(n;1)*1*x+C(n;2)*1*x²+...+C(n;n-1)*x^(n-1) +C(n;n)*x^n
n(1+x)^(n-1)=C(n;1)+2C(n;2)*x+3*C(n;3)x²+...+(n-1)*C(n;n-1)*x^(n-2)+n*C(n;n)*x^(n-1)
x=1
C(n;1)+2*C(n;2)+3*C(n;3)+...+n*C(n;n)=n*2^(n-1)
Author:
Math Problems
#Problems #Quest6
#Solutions | #Quest6
So here there are only 7 people, where the 7th person is both a lawyer and CA. We will split the problem into cases: when the "both" person is part of the committee and when he isn't.
Case 1: For this committee, the "both" person must be part of it, so we really are choosing 2 people from the remaining 6. So we have 1⋅C6,2
Case 2: Here, the "both" person isn't part of the committee, so we have 3 Lawyers we can choose for our necessary lawyer, 3 for our necessary CA, and finally 4 for the remaining person who can be anything. This leaves us with 3⋅3⋅4 committees. However, we don't care about distinguishing between when the person is chosen, so we divide by the permutations of all committees, leaving us with 3⋅3⋅4/3!=6.
Adding the two cases, we get 21.
Author:@UltimateTOE
So here there are only 7 people, where the 7th person is both a lawyer and CA. We will split the problem into cases: when the "both" person is part of the committee and when he isn't.
Case 1: For this committee, the "both" person must be part of it, so we really are choosing 2 people from the remaining 6. So we have 1⋅C6,2
Case 2: Here, the "both" person isn't part of the committee, so we have 3 Lawyers we can choose for our necessary lawyer, 3 for our necessary CA, and finally 4 for the remaining person who can be anything. This leaves us with 3⋅3⋅4 committees. However, we don't care about distinguishing between when the person is chosen, so we divide by the permutations of all committees, leaving us with 3⋅3⋅4/3!=6.
Adding the two cases, we get 21.
Author:
Math Problems
#Problem #Quest44
#Solutions | #Quest44
This is basically the number of ways to select a committee with at least k people, have k people be the leaders of the committee, then choose a person out of the other people not in the committee(but in the n people). Therefore, we can instead just choose k people as the leaders first, then the person out of the other people, then everyone else can be in or not in the committee, so the answer is binom (n)(k) . (n-k). 2^(n-k-1).
Author: @MathMrIO
This is basically the number of ways to select a committee with at least k people, have k people be the leaders of the committee, then choose a person out of the other people not in the committee(but in the n people). Therefore, we can instead just choose k people as the leaders first, then the person out of the other people, then everyone else can be in or not in the committee, so the answer is binom (n)(k) . (n-k). 2^(n-k-1).
Author: @MathMrIO
Math Problems
#Problems #Quest76
#Solutions | #Quest76
deg(P(x))=1 => P(x)=kx+c
For all x€R P(x) increasing.
Therefore max(P(x))=kb+c
min(P(x))=ka+c
maxP(x)-minP(x)=k(b-a) (1)
we have
maxP(x)-minP(x)=b-a (2)
From (1) and (2)
k(b-a)=b-a => k=1
P(x)=kx+c=x+c
c€R.
Answer: x+c , c€R
Author:Jamshidbek Hasanov
deg(P(x))=1 => P(x)=kx+c
For all x€R P(x) increasing.
Therefore max(P(x))=kb+c
min(P(x))=ka+c
maxP(x)-minP(x)=k(b-a) (1)
we have
maxP(x)-minP(x)=b-a (2)
From (1) and (2)
k(b-a)=b-a => k=1
P(x)=kx+c=x+c
c€R.
Answer: x+c , c€R
Author: