Question

The total number of ways in which a beggar can be given at least one rupee from four $25$ paise coins and three $50$ paise coins and $2$ one rupee coins is
A. $54$
B. $50$
C. $52$
D. None

Answer 1

Hint: First try to find out total number of ways beggar get zero rupees or maximum rupees so that is equal to the $\underbrace {4 + 1}_{{\text{either give all 25paisa coin or didn't give any}}}$
similarly for three $50$ paise coins and $2$ one rupee coins Total number of ways = \[\left( {4 + 1} \right).(3 + 1).(2 + 1)\]=$60$
Now remove the number of ways that he gets less than $1$ rupee that are $6$ ways in which he gets to subtract it and get the answer .

Complete step-by-step answer:
In this question we have to give to the beggar at least one rupee from four $25$ paise coins and three $50$ paise coins and $2$ one rupee coins so for this first find the total number of ways that beggar can earn zero rupees or maximum rupees for this ,
We have four $25$ paise coins.
We have three $50$ paise coins.
We have $2$ one rupee coins.
So total number of cases that beggar get zero rupees or maximum rupees is,
From four $25$ paise coins either we will give all the four coins or we didn't give any number of coins so the number ways is $4 + 1$
Similarly,
From three $50$ paise coins either we will give all the three coins or we didn't give any number of coins so the number ways is $3 + 1$
From $2$ one rupees coins either we will give all the two coins or we didn't give any number of coins so the number ways is $2 + 1$
 Total number of ways = \[\left( {4 + 1} \right).(3 + 1).(2 + 1)\]
= $5.4.3$
=$60$ ways by which beggars get zero rupees or maximum rupees .
Now ,
we have to find the number of beggars that beggars can get at least $1$ rupee .
\[ \Rightarrow \]number of ways that beggar get $0$ rupees is $1$
\[ \Rightarrow \]number of ways that beggar get $25$ paisa is $1$
\[ \Rightarrow \]number of ways that beggar get $50$ paisa is $2$ as we give a beggar two $25$ paisa coin or one $50$ paisa coin
\[ \Rightarrow \]number of ways that beggar get $75$ paisa is $2$ as we give a beggar three $25$ paisa coin or one $50$ paisa coin and one $25$ paisa coin
\[ \Rightarrow \]Number of ways that beggar will get less than $1$ rupees is $1 + 1 + 2 + 2 = 6$ ways
hence ,
A beggar can be given at least one rupee is =
 number of cases that beggar get zero rupees or maximum rupee - Number of ways that beggar will get less than $1$ rupees
\[ \Rightarrow \]$60 - 6$
$ = 54$ ways, the beggar can be given at least one rupee .

Hence, option A is correct.

Note: The number of selections from n different objects, taking at least one = \[{\;^n}{C_1}\; + {\;^n}{C_2}\; + {\;^n}{C_3}\; + {\text{ }}...{\text{ }} + {\;^n}{C_n}\; = {\text{ }}{2^n}\; - {\text{ }}1\]
If we have to arrange anything like people, digits, numbers, alphabets, letters, and colours then we have to use permutation.
If we have to Selection of menu, food, clothes, subjects, and team then we have to combine formulas.