Four friends were walking with a bag full of identical coins. They fo

Question

Four friends were walking with a bag full of identical coins. They found a beggar and decided to given him some coins. They formulated the concept that each one of them will go to the beggar separately with as many number of coins as one wishes but the number should not be zero nor it should exceed 5. How many ways there can be for the four friends to give the beggar some coins?

A. 5!

B. \(5^4\)

C. 5P4

D. \(\frac{6!}{5!}\)

E. 4*5!

A. 5!

B.  5^4

C. 5P4

D.  \frac{6!}{5!}

E. 4*5!

Lampard42 · Answer

Every person has five possibilities and 4 persons. So each has 5 options. So 5^4 IMO B

CareerGeek · Answer

Number should not be zero nor it should exceed 5 
--> Each person can give any number from 1 to 5 coins
--> Each person has 5 options
--> 4 persons will have 5*5*5*5 = 5^4 (Repetition is allowed as 2 or more persons can give same number of coins)

Option B

Pls Hit kudos if you like the solution

EgmatQuantExpert · Answer

Solution

Given: 
 • Total number of friends = 4
• Each one of them will go to the beggar separately with as much number of coins as one wish, but the number should not be zero nor should exceed 5

To Find: 
 • The number of ways there can be, for the four friends to give the beggar some coins

Approach & Working Out: 
 • The number of coins each one can give the beggar is one among (1, 2, 3, 4, 5)
• Thus, the number of ways in which each one can give some coins = 5
• Therefore, the four friends can give the beggar some coins in  5 * 5 * 5 * 5 = 5^4  ways

Hence, the correct answer is  Option B

ScottTargetTestPrep · Answer

Since each person can give any number from 1 to 5 coins to the beggar, each has 5 options. Therefore, the number of ways of giving coins to the beggar is 5 x 5 x 5 x 5 = 5^4.

Answer: B

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