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 ar
1 answer:
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
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Answer:
Step-by-step explanation:
Easy way to do this is step by step. Your quadratic, from your entry, must be
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Step by step looks like this, one thing at a time:
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Do the same with the subtraction sign to get the other solution.
If you're unsure of how to enter it into your calculator, do it step by step so you don't mess up the sign. If you enter it incorrectly, you could end up with an imaginary number when it should be real, or a real one that should be imaginary.
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