In how many ways can people be lined up to get on a bus if a particular person always boards the bus first or last?
1 answer:
Incomplete question. I Assumed the number of people is 6. In other words, In how many ways can 6 people be lined up to get on a bus if a particular person always boards the bus first or last?
Answer:
<u>6</u>
Step-by-step explanation:
Note, this question involves the use of the permutation formula; which helps <u>determine the arrangement or rearrangement of a set of objects in an ordered way.</u> Since we are told a particular person (ie 1 person) from among the group always boards first or last, it forms our r.
P(n,r)= where n = number of people; r = the difference taken at a time.
P(n,r) = =
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The simplified expression is as follows:
d(-3x^2 + 9x - 14)
Answer:
x = 2 , y = 0 , z = 3
Step-by-step explanation:
Cramer's rule is a rule through which we can find the solution of linear equation.
we have the three linear equations as
x+2y+3z=11
2x+y+2z=10
3x+2y+z=9
AX=B
A: coefficient matrix
X= unknown vectors(x,y,z)
D = values of the linear equation (11 , 10 , 9)
now we find the determinant of the given linear equation
determinant of the matrix will be
A = = 1(1-4) - 2(2-6) + 3(4 - 3)
= 1(-3) - 2(-4) + 3(1)
= -3+8+3 = 8
also D
so the determinant is Non zero we can apply Cramer's rule
we will be replacing the first column of the coefficient matrix A with the values of D
by replacing the first column we will get the value of the variable 'x'
Dx = = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16
x = =
= 2
similarly
Dy = = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0
y = = 0
Dz= = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24
z = =
so we have the solution as
x = 2 , y = 0 , z = 3
Therefore the solution for the given linear equations is (2,0,3).