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Martinez Warren
12/01/2023 · Middle School
(a) \(57120\)
(b) \(2380\)
UpStudy Free Solution:
Let's solve the problem of choosing 4 objects from 17 distinct objects in two scenarios: when the order of choices matters and when it doesn't.
(a) Order of Choices Matters
When the order of selection matters, we are dealing with permutations. The number of ways to choose 4 objects out of 17, taking the order into consideration, is given by the permutation formula:
\[P( n, k) = \frac { n! } { ( n - k) ! } \]
Here, \(n = 17\) and \(k = 4\):
\[P( 17, 4) = \frac { 17! } { ( 17 - 4) ! } = \frac { 17! } { 13! } \]
This simplifies to:
\[P( 17, 4) = 17 \times 16 \times 15 \times 14\]
Calculating this:
\[17 \times 16 = 272\]
\[272 \times 15 = 4080\]
\[4080 \times 14 = 57120\]
So, the number of ways to choose 4 objects out of 17, considering the order, is:
\[57120\]
(b) Order of Choices Does Not Matter
When the order of selection does not matter, we are dealing with combinations. The number of ways to choose 4 objects out of 17, without considering the order, is given by the combination formula:
\[C( n, k) = \binom { n} { k} = \frac { n! } { k! ( n - k) ! } \]
Here, \(n = 17\) and \(k = 4\):
\[C( 17, 4) = \binom { 17} { 4} = \frac { 17! } { 4! ( 17 - 4) ! } = \frac { 17! } { 4! \cdot 13! } \]
This simplifies to:
\[\binom { 17} { 4} = \frac { 17 \times 16 \times 15 \times 14} { 4 \times 3 \times 2 \times 1} \]
Calculating this:
\[17 \times 16 = 272\]
\[272 \times 15 = 4080\]
\[4080 \times 14 = 57120\]
\[4 \times 3 = 12\]
\[12 \times 2 = 24\]
\[24 \times 1 = 24\]
Now, divide:
\[\frac { 57120} { 24} = 2380\]
So, the number of ways to choose 4 objects out of 17, without considering the order, is:
\[2380\]
Summary
(a) If the order of the choices is taken into consideration, there are \(57120\) ways to choose 4 objects out of 17.
(b) If the order of the choices is not taken into consideration, there are \(2380\) ways to choose 4 objects out of 17.
Supplemental Knowledge
In combinatorics, choosing objects can be approached in two ways: permutations and combinations.
Permutations consider the order of selection. The number of permutations of \(n\) objects taken \(r\) at a time is given by:
\[P( n, r) = \frac { n! } { ( n- r) ! } \]
Combinations do not consider the order of selection. The number of combinations of \(n\) objects taken \(r\) at a time is given by:
\[C( n, r) = \binom { n} { r} = \frac { n! } { r! ( n- r) ! } \]
Understanding permutations and combinations can be daunting without proper guidance, which is where UpStudy comes in!
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