Question

In a cloth showroom there are 4 kinds of belts, three kinds of ties, two kinds of shoes and seven kinds of dresses. In how many ways can a person decorate himself if he wants to select only one belt, one tie and one of the following a shoe or a dress

Answer 1

Answer:

This is a question based on Combinations.

Given,

• 4 kinds of belt
• 3 kinds of tie
• 2 kinds of shoe
• 7 kinds of dress

The person has to select:

• 1 belt out of 4.
• 1 tie out of 3.
• 1 pair of shoe fout of 2 (or) 1 kind of dress out of 7

The number of combinations in which 1 belt can be chosen from 4 different kinds is:

$\implies ^4C_1 = \dfrac{4!}{(4-1)!.\:1!}\\\\\\\implies ^4C_1 = \dfrac{4!}{3!.\:1!}\\\\\\\implies ^4C_1 = \dfrac{ 4 \times 3!}{ 3!. \: 1!}\\\\\\\implies ^4C_1 = \dfrac{4}{1} = \boxed{ \bf{ 4 }}$

The number of combinations in which 1 tie can be chosen from 3 different kinds is:

$\implies ^3C_1 = \dfrac{3!}{(3-1)!.\: 1!}\\\\\\\implies ^3C_1 = \dfrac{3!}{2!.\: 1!} \\\\\\\implies ^3C_1 = \dfrac{3 \times 2!}{2!. \: 1! }\\\\\\\implies ^3C_1 = \dfrac{3}{1} = \boxed{ \bf{ 3 }}$

The number of combinations in which 1 pair of shoes can be selected from 2 different kinds is:

$\implies ^2C_1 = \dfrac{ 2! }{ (2-1)!. \: 1! }\\\\\\\implies ^2C_1 = \dfrac{ 2 \times 1! }{ 1!. \: 1!}\\\\\\\implies ^2C_1 = \dfrac{ 2 }{ 1 } = \boxed{ \bf{ 2 }}$

The number of combinations in which 1 set of dress can be selected from 7 different kinds is:

$\implies ^7C_1 = \dfrac{ 7! }{ (7-1)! . \: 1! }\\\\\\\implies ^7C_1 = \dfrac{ 7 \times 6! }{ 6!. \: 1!}\\\\\\\implies ^7C_1 = \dfrac{7}{1!} = \dfrac{7}{1} = \boxed{ \bf{ 7 }}$

Since the belt and the tie has no choice of selection in between them, we multiply their combinations. But on the other hand, we have choice "(or)" between shoes and dress. Hence we add their combinations.

Therefore the total number of combinations is:

⇒ 4 × 3 × ( 2 + 7 )

⇒ 12 × 9

⇒ 108 combinations

Hence the number of ways in which a person can decorate himself is 108.

Answer 2

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{• \: 4 \: kinds \: of \: belts} \\ &\sf{• \: 3 \: kinds \: of \: ties}\\ &\sf{• \: 2 \: kinds \: of \: shoes}\\ &\sf{• \: 7 \: kinds \: of \: dress} \end{cases}\end{gathered}\end{gathered}$

$\large\underline\blue{\bold{To \:  Find :-  }}$

$\begin{gathered}\begin{gathered}\tt number \: of \: ways \: of \: selecting= \begin{cases} &\sf{• \: one \: belt} \\ &\sf{• \: one \: tie} \\ &\sf{• \: one \: dress \: or \: one \: shoe}\end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\tt \: Concept \: used :- \begin{cases} &\sf{combinations} \end{cases}\end{gathered}\end{gathered}$

If there are 'n' distinct objects, out of which 'r' objects have to be chosen, then number of ways of this selection is

$\begin{gathered}\implies ^nC_r = \dfrac{n!}{(n-r)!.\: r!}\end{gathered}$

☆ Formula Used:-

Formula used :- $\large \boxed{\tt \:  ⟼ \: ^nC_1 \: = n }$

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Solution:-

$\tt \:  ⟼ \: Total \: number \: of \: belts = 4$

$\tt \:  ⟼ \: Number \: of \: belts \: to \: be \: selected = 1$

$\boxed{\tt \:  So, \: number \: of \: ways \: to \: select \: belt = ^4C_1 = 4}$

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$\tt \:  ⟼Total \: number \: of \: ties = 3$

$\tt \:  ⟼Number \: of \: ties \: to \: be \: selected = 1$

$\boxed{\tt \: So, \: number \: of \: ways \: to \: select \: 1 \: tie= ^3C_1 \: = 3 }$

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☆ Now he has to choose only one thing out of 7 dresses or 2 shoes.

$\boxed { \: \tt \: So, \: number \: of ways \: of \: selection = \: ^9C_1 \: = 9 }$

Hence, the total number of selecting one tie, one belt and one dress or one shoe is

$\boxed{\tt\implies \:number \: of \: ways = 4 \times 3 \times 9 = 108 \: ways}$