1)SOLVE:
a) 2nP3 = 2(nP4)
b)6(n+1P2) = 3nP2
2) If (nC12 = (nC8), find (nC17) and (22Cn)
3)simplify.
a) n(squared) (n-2)! - n(n-2)! / n!
b) (nC2) - (nCn-2)
4)If nPr - 506 and (nCr_ = 253, find n and r.
5) If (28C2r)/(24C2r-4) = 225/11, find r.
a)
P(2n,3)=2P(n,4)
=>
2n(2n-1)(2n-2)=2n(n-1)(n-2)(n-3)
cancel 2n to get
(2n-1)(2n-2)=(n-1)(n-2)(n-3)
This kind of equation can be readily solved by trial and error, since they both increase monotonically at different rates.
In this case, n=1 (which is rejected) or n=8.
try (b) and (2)similarly to (a) above.
3(a) simplifies well, assuming you have left out the critical square brackets:
[n²(n-2)!-n(n-2)!]/n!
(n-2)![n²-n]/n!
=(n-2)!n[n-1]/n!
=n!/n!
=1
Give a try to 3b.
4.
What it is saying is that
P(n,r)=2C(n,r)
=>
n!/(n-r)! = 2*n!/((n-r)!r!)
Cancel the n! and (n-r)! to get
1=2/r!
=> r!=2 => r=2
After that, you only have to check by trial and error C(n,2)=253.
Give (5) a try.
np4=88x n-1 p2
the question is very important question but you may answer each question step by step please
1)
a) To solve 2nP3 = 2(nP4), we can start by expanding the permutations using the factorial notation.
nP3 = n! / (n-3)! and nP4 = n! / (n-4)!
Substituting these expressions into the equation, we get:
2(n! / (n-3)!) = 2(n! / (n-4)!)
We can cancel out the common terms on both sides:
n! / (n-3)! = n! / (n-4)!
Now we have:
1 / (n-3)! = 1 / (n-4)!
Since the denominators are equal, the numerators must also be equal:
n-3 = n-4
Simplifying this equation, we find:
-3 = -4
This equation is not possible, so there is no solution to this equation.
b) For 6(n+1P2) = 3nP2, we follow a similar approach:
n+1P2 = (n+1)! / (n+1-2)! = (n+1)! / n!
nP2 = n! / (n-2)!
Substituting these expressions into the equation:
6((n+1)! / n!) = 3(n! / (n-2)!)
Now we can cancel out the common terms:
6(n+1)! / n! = 3(n!) / (n-2)!
Cross multiplying:
6(n+1)! * (n-2)! = 3(n!) * n!
Simplifying:
6(n+1)(n)(n-1)(n-2)! = 3(n)(n-1)(n-2)!
Dividing both sides by 3(n)(n-1)(n-2)!, we get:
2(n+1) = 1
n + 1 = 1/2
Subtracting 1 from both sides:
n = -1/2
Therefore, the solution to the equation is n = -1/2.
2)
Given (nC12) = (nC8), we need to find (nC17) and (22Cn).
We know that (nCr) = n! / [r!(n-r)!].
So, (nC12) = n! / [12!(n-12)!],
and (nC8) = n! / [8!(n-8)!].
Setting these two equations equal, we have:
n! / [12!(n-12)!] = n! / [8!(n-8)!].
By canceling out the common terms, we get:
1 / [12!(n-12)!] = 1 / [8!(n-8)!].
Simplifying further:
1 / [12! * (n-12)!] = 1 / [8! * (n-8)!].
Since the denominators are equal, the numerators must also be equal:
12! * (n-12)! = 8! * (n-8)!
Now, we can solve for (nC17) and (22Cn) in a similar manner.
For (nC17), we use the formula:
(nC17) = n! / [17!(n-17)!],
and for (22Cn), we use:
(22Cn) = 22! / [n!(22-n)!].
3)
a) To simplify n^2 * (n-2)! - n * (n-2)! / n!,
we can factor out the common term (n-2)!:
(n-2)! * (n^2 - n) / n!.
Simplification is not possible without knowing the specific value of n.
b) To simplify (nC2) - (nCn-2),
we can calculate the binomial coefficients:
(nC2) = n! / [2!(n-2)!] = n(n-1) / 2,
(nCn-2) = n! / [(n-2)!(n-(n-2))!] = n(n-1) / 2.
Substituting these values into the equation, we get:
(nC2) - (nCn-2) = n(n-1) / 2 - n(n-1) / 2.
Since the two terms have the same denominator, we can combine them:
[n(n-1) - n(n-1)] / 2 = 0.
Therefore, the simplified form of the expression is 0.
4)
Given nPr = 506 and (nCr) = 253, we need to find the values of n and r.
Using the formulas for permutations and combinations:
nPr = n! / (n-r)! = 506,
(nCr) = n! / [r!(n-r)!] = 253.
We can rearrange the nPr equation to solve for n-r:
n-r = n - nPr.
Substituting the values given:
n - nPr = n - 506.
This implies that nPr = 506.
Now, we can substitute these values into the (nCr) equation:
n! / [r!(n-r)!] = 253.
We can substitute nPr from the previous equation:
(n+r)! / [r!(n-r)!] = 253.
Multiplying both sides by [r!(n-r)!] to simplify:
(n+r)! = 253r!(n-r)!.
Since we know that nPr = 506, we can substitute this value:
(n-506)! = 253r!(506-r)!.
We can simplify further by dividing both sides by 253:
(n-506)! / 253 = r!(506-r)!.
It is not possible to determine the exact values of n and r without additional information or simplification.
5)
Given (28C2r) / (24C2r-4) = 225/11, we need to find the value of r.
From the formulas for combinations:
(28C2r) = 28! / [2r!(28-2r)!],
(24C2r-4) = 24! / [2(r-2)!(24-2(r-2))!].
Substituting these values into the equation, we have:
[28! / (2r!(28-2r)!)] / [24! / [2(r-2)!(24-2(r-2))!]] = 225/11.
We can simplify by dividing both sides by 1/[(2r!(28-2r)!)(2(r-2)!(24-2(r-2))!)], which is equivalent to multiplying by the reciprocal:
([28! / (2r!(28-2r)!)] / [24! / [2(r-2)!(24-2(r-2))!]]) * [(2(r-2)!(24-2(r-2))!) / 1] = (225/11) * [(2r!(28-2r)!)(2(r-2)!(24-2(r-2))!)].
Simplifying further, we get:
[28! / 24!] * [1 / (2r!(28-2r)!)] * [1 / (2(r-2)!(24-2(r-2))!)] = (225/11).
Cancelling out the common terms, we have:
[(28*27*26*25) / (24*23*22*21)] * [1 / (2r!(28-2r)!)] * [1 / (2(r-2)!(24-2(r-2))!)] = (225/11).
Simplifying both sides, it is not possible to determine the exact value of r without additional information or simplification.