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Question: There are \(10\) seats in a double-decker bus, \(6\) in the lower deck, and \(4\) on the upper deck....

There are 1010 seats in a double-decker bus, 66 in the lower deck, and 44 on the upper deck. Ten passengers board the bus, of them 33 refuse to go to the upper deck and 22 insist on going up. The number of ways in which the passengers can be accommodated is______. (Assume all seats to be duly numbered)

Explanation

Solution

There are ten seats in the bus and ten passengers board the bus in which 33 refuse to go up and 22 insist on going up. So they have to be given the seats according to their choice but the rest of the 55 passengers can be seated in any order. So we will use the formula of combination to solve this question which is nCr=n!r! n - r!{}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r! n - r!}}}}

Complete step by step answer:

Given, the total number of seats is 1010.The number of seats in the lower deck is 66and in the upper deck is 44.The number of passengers boarding the bus is also 1010.Now out of 10 passengers, 33 refuses to go up and 22 insists on going up. We know that the number of ways to select r things from n things =nCr = {}^{\text{n}}{{\text{C}}_{\text{r}}} and the number of ways the n seats can be filled is (n!). So the number of ways in which 2 out of 4 upper deck seats are selected and filled by the 22 people who insist on going up =4C2×2! = {}^4{{\text{C}}_2} \times 2!
Similarly the number of ways in which the 3 out of 6 seats can be selected for the people, who refuse to go up, and filled by the 3 people =6C3×3! = {}^6{{\text{C}}_3} \times 3!. In this way 5 seats are filled, now only 5 seats and 5 people are left. So the number of ways in which 55 seats can be filled by 55people=5! = 5!
So the number of ways in which the passengers can be accommodated =4C2×2!×6C3×3!×5! = {}^4{{\text{C}}_2} \times 2! \times {}^6{{\text{C}}_3} \times 3! \times 5!
The formula of combination nCr=n!r! n - r!{}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r! n - r!}}}} where n is the total number of elements in a set and r are the number of elements to be selected, (!) is the sign of factorial. Here,n!=n×(n - 1)×...×3×2×1=n×(n−1)!{\text{n}}! = {\text{n}} \times \left( {{\text{n - 1}}} \right) \times ... \times 3 \times 2 \times 1 = {\text{n}} \times \left( {{\text{n}} - 1} \right)!
On putting the given value in the formula we get-
The number of ways in which the passengers can be accommodated =4!2!4−2!×2!×6!3!6−3!×3!×5!=4!2!×6!3!×5! = \dfrac{{4!}}{{2!4 - 2!}} \times 2! \times \dfrac{{6!}}{{3!6 - 3!}} \times 3! \times 5! = \dfrac{{4!}}{{2!}} \times \dfrac{{6!}}{{3!}} \times 5!
On simplifying we get,
⇒4×3×2!2!×6×5×4×3!3!×5!=12×120×120=172,800\Rightarrow \dfrac{{4 \times 3 \times 2!}}{{2!}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!}} \times 5! = 12 \times 120 \times 120 = 172,800
Answer-The number of ways the passengers can be accommodated is 172,800172,800.

Note: Combinations can be confused with permutations. However, in permutations, the order of the selected items is essential. A combination determines the number of possible arrangements in a collection of items where the order of the selection does not matter. The items can be selected in any order.