Question

How many ways are there to choose 6 items from 10 distinct items when

a) the items in the choices are ordered and repetition is not allowed?

b) the items in the choices are ordered and repetition is allowed?

c) the items in the choices are unordered and repetition is not allowed?

d) the items in the choices are unordered and repetition is allowed?

Short answer

(a) The number of ways to choose 6 items from 10 distinct items when the items in the choices are ordered and repetition is not allowed is 151,200 ways.

(b) The number of ways to choose 6 items from 10 distinct items when the items in the choices are ordered and repetition is allowed is 1,000,000 ways.

(c) The number of ways to choose 6 items from 10 distinct items when the items in the choices are unordered and repetition is not allowed is 210 ways.

(d) The number of ways to choose 6 items from 10 distinct items when the items in the choices are unordered and repetition is allowed is 5005 ways.

Step-by-step solution

Concept Introduction

Definition of permutation (order is important) is –

No repetition allowed:\({\rm{P(n,r) = }}\frac{{{\rm{n!}}}}{{{\rm{(n - r)!}}}}\)

Repetition allowed:\({{\rm{n}}^{\rm{r}}}\)

Definition of combination (order is important) is –

No repetition allowed:\({\rm{C(n,r) = }}\frac{{{\rm{n!}}}}{{{\rm{r!(n - r)!}}}}\)

Repetition allowed:\({\rm{C(n + r - 1,r) = }}\frac{{{\rm{(n + r - 1)!}}}}{{{\rm{r!(n - 1)!}}}}\)

With\({\rm{n! = n}} \cdot {\rm{(n - 1)}} \cdot ... \cdot {\rm{2}} \cdot {\rm{1}}\).

Calculation for choices that are ordered and repetition is not allowed

(a)

Order is important (choices are ordered), thus it is needed to use permutation.

Here, it can be seen\({\rm{n = 10, r = 6}}\).

Since repetition is not allowed, so substitute the value and calculate –

\(\begin{array}{c}{\rm{P(10,6) = }}\frac{{{\rm{10!}}}}{{{\rm{(10 - 6)!}}}}\\{\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{4!}}}}\\{\rm{ = 10}} \cdot {\rm{9}} \cdot {\rm{8}} \cdot {\rm{7}} \cdot {\rm{6}} \cdot {\rm{5}}\\{\rm{ = 151,200}}\end{array}\)

Therefore, the result is obtained as \({\rm{151,200}}\).

Calculation for choices that are ordered and repetition is allowed

(b)

Order is important (choices are ordered), thus it is needed to use permutation.

Here, it can be seen\({\rm{n = 10, r = 6}}\).

Since repetition is allowed, so substitute the value and calculate –

\(\begin{array}{c}{{\rm{n}}^{\rm{r}}}{\rm{ = 1}}{{\rm{0}}^{\rm{6}}}\\{\rm{ = 1,000,000}}\end{array}\)

Therefore, the result is obtained as \({\rm{1,000,000}}\).

Calculation for choices that are unordered and repetition is not allowed

(c)

Order is not important (choices are unordered), thus it is needed to use combination.

Here, it can be seen\({\rm{n = 10, r = 6}}\).

Since repetition is not allowed, so substitute the value and calculate –

\(\begin{array}{c}{\rm{C(10,6) = }}\frac{{{\rm{10!}}}}{{{\rm{6!(10 - 6)!}}}}\\{\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{6!4!}}}}\\{\rm{ = }}210\end{array}\)

Therefore, the result is obtained as \({\rm{210}}\).

Calculation for choices that are unordered and repetition is allowed

(d)

Order is not important (choices are unordered), thus it is needed to use combination.

Here, it can be seen\({\rm{n = 10, r = 6}}\).

Since repetition is allowed, so substitute the value and calculate –

\(\begin{array}{c}{\rm{C(n + r - 1,r) = C(10 + 6 - 1,6)}}\\{\rm{ = C(15,6)}}\\{\rm{ = }}\frac{{{\rm{15!}}}}{{{\rm{6!(15 - 6)!}}}}\\{\rm{ = }}\frac{{{\rm{15!}}}}{{{\rm{6!9!}}}}\\{\rm{ = 5005}}\end{array}\)

Therefore, the result is obtained as \({\rm{5005}}\).