2x - y = 64x-y13anch
2 answers:
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This question is incomplete, the complete question is;
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.
In other words, how many 5-tuples of integers ( h, i , j , m ), are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?
Answer:
the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Step-by-step explanation:
Given the data in the question;
Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1
this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.
So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'
Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;
= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]
= [( n + 4 )!] / [ 5!( n-1 )! ]
= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120
Answer:
Step-by-step explanation:
From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e () and are parallel to
The parallel vector to the line i + zj+k = ai + bj + ck
Hence, the equation for the line is :
x = 4 + t
y = -5 + 2t
z = 2 + t
Thus, x, y, z = ( 4+t, -5+2t, 2+t )
The symmetric equation can now be as follows:
∴