Passengers are to travel by a double decked bus which can accommodate 13 in the upper deck and 7 in the lower deck. The number of ways that they can be distributed if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck is:
1. 25
2. 21
3. 18
4. 15
Answer
536.4k+ views
Hint: For solving this question you should know about permutations and combinations. As we know that this problem has statements only and we have to tell whether these are true or false. So, first we will discuss the permutations and combinations and then by applying the concept of these we will find the solution of the problem and then will find the existence of the given statement. The formula of combination is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Complete step-by-step solution:
According to the question it is asked that there are 13 seats in upper decks and 7 seats in the lower deck. We have to determine the number of ways that they can be distributed if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck. As we know, permutations are ordered combinations. It means where the order does matter there is a permutation. And permutations have also two types, one is repeated or repetition is allowed and the second is no repetition. The permutations with repetition are the easiest to calculate. And the permutation with no repetition reduces the number of available choices each time.
Therefore 5 have refused for the upper deck. So they will be in the lower deck and 8 will be in the upper deck.
So, the remaining people are = 7
Seats remaining in upper deck = 5
Number of ways = $^{7}{{C}_{5}}$
And people remaining = 2 and the seats are in the lower deck = 2.
Therefore the number of ways = $^{2}{{C}_{2}}$
Therefore total number of ways = $^{7}{{C}_{5}}{{\times }^{2}}{{C}_{2}}=21$
Hence the total number of ways = 21.
So, the correct option is 2.
Note: While solving this type of questions you have to ensure that we will use permutation or combination here and in the permutation it is necessary to find if there are any events that are repeating or not repeating.
Complete step-by-step solution:
According to the question it is asked that there are 13 seats in upper decks and 7 seats in the lower deck. We have to determine the number of ways that they can be distributed if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck. As we know, permutations are ordered combinations. It means where the order does matter there is a permutation. And permutations have also two types, one is repeated or repetition is allowed and the second is no repetition. The permutations with repetition are the easiest to calculate. And the permutation with no repetition reduces the number of available choices each time.
Therefore 5 have refused for the upper deck. So they will be in the lower deck and 8 will be in the upper deck.
So, the remaining people are = 7
Seats remaining in upper deck = 5
Number of ways = $^{7}{{C}_{5}}$
And people remaining = 2 and the seats are in the lower deck = 2.
Therefore the number of ways = $^{2}{{C}_{2}}$
Therefore total number of ways = $^{7}{{C}_{5}}{{\times }^{2}}{{C}_{2}}=21$
Hence the total number of ways = 21.
So, the correct option is 2.
Note: While solving this type of questions you have to ensure that we will use permutation or combination here and in the permutation it is necessary to find if there are any events that are repeating or not repeating.
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