There are 10 lamps in a hall. Each one of them can be switched on independently. find the number of ways in which the hall can be illuminated.
We have to find the possible number of ways in which we can illuminate the hall by lighting the lamps which are independent of each other.
We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.
We will use a combination technique in this question, which will make us solve the question in an easy way. The combination means selection in rough language.
The combination is denoted by the symbol νCr , which means a number of ways of selecting r objects at a time from n objects.
We will make cases by choosing a specific number of lamps at one time,
Taking one lamp at a time, we can enlighten the hall in 10C1 ways.
Taking two lamps at a time, we can enlighten the hall in 10C2 ways.
Taking three lamps at a time, we can enlighten the hall in 10C3 ways
Taking four lamps at a time, we can enlighten the hall in 10C4 ways.
Taking five lamps at a time, we can enlighten the hall in 10C5 ways.
Taking six lamps at a time, we can enlighten the hall in 10C6 ways.
Taking seven lamps at a time, we can enlighten the hall in 10C7 ways.
Taking eight lamps at a time, we can enlighten the hall in 10C8 ways.
Taking nine lamps at a time, we can enlighten the hall in10C9 ways.
Taking ten lamps at a time, we can enlighten the hall in10C10 ways.
By using the binomial expansion (1+x)10 , if we put x = 1, then we will get the sum of these binomial coefficients in which there is extra 10C0 term which is equal to 1, so we subtract 1 from the sum.
Therefore the total number of ways of enlightening the hall are = 210 – 1