
In a 12 storey house, 9 people enter in a lift cabin. It is known that they will leave the lift in a group of 2, 3 and 4 people at different storeys. In how many ways can they do so if the lift does not stop at the second storey?
Answer
483.6k+ views
Hint:
In this problem they have asked us the number of ways that event can happen so we will use combinations. But the condition is the lift is not stopping at the second floor so that floor would not be considered in selection. Thus the number of possible floors on which lift will stop is 11.One more thing is the people are already leaving the lift in a group of 2, 3 and 4. So we need not make any change. Also it says that we can’t judge which group will leave the lift first. So they also will have combinations. Hope we are clear with the question. Let’s solve it!
Complete step by step solution:
It is a 12 storey building. But the lift will not stop on the second floor. So the total number of floors in consideration are 11.
But there are only three floors that will be used from these 11. So the total combinations will be \[11{C_3}\]
Now these groups will leave the lift in any sequence. So that combination is also considered. It is \[3!\].
So, total number of ways this event can take place are
\[ \Rightarrow {}^{11}{C_3} \times 3!\]
\[ \Rightarrow \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} \times 3!\]
\[ \Rightarrow \dfrac{{10!}}{{3!7!}} \times 3!\]
Cancel that 3! term.
\[ \Rightarrow \dfrac{{10!}}{{7!}}\]
\[ \Rightarrow 10 \times 9 \times 8\]
\[ \Rightarrow 720\]
So these are the total number of ways in which the groups of people will leave the lift cabin.
Note:
In this problem students can miss the possible ways of groups of people that change. Because we are not given the sequence in which the groups will leave. So that also needs to be considered.
One more note is that in question they are given the lift will not stop at $2^{\text{nd}}$ floor that’s why only that floor is excluded. But if the words are like upto $2^{\text{nd}}$ floor then we will exclude 2 floors $1^{\text{st}}$ and $2^{\text{nd}}$.
In this problem they have asked us the number of ways that event can happen so we will use combinations. But the condition is the lift is not stopping at the second floor so that floor would not be considered in selection. Thus the number of possible floors on which lift will stop is 11.One more thing is the people are already leaving the lift in a group of 2, 3 and 4. So we need not make any change. Also it says that we can’t judge which group will leave the lift first. So they also will have combinations. Hope we are clear with the question. Let’s solve it!
Complete step by step solution:
It is a 12 storey building. But the lift will not stop on the second floor. So the total number of floors in consideration are 11.
But there are only three floors that will be used from these 11. So the total combinations will be \[11{C_3}\]
Now these groups will leave the lift in any sequence. So that combination is also considered. It is \[3!\].
So, total number of ways this event can take place are
\[ \Rightarrow {}^{11}{C_3} \times 3!\]
\[ \Rightarrow \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} \times 3!\]
\[ \Rightarrow \dfrac{{10!}}{{3!7!}} \times 3!\]
Cancel that 3! term.
\[ \Rightarrow \dfrac{{10!}}{{7!}}\]
\[ \Rightarrow 10 \times 9 \times 8\]
\[ \Rightarrow 720\]
So these are the total number of ways in which the groups of people will leave the lift cabin.
Note:
In this problem students can miss the possible ways of groups of people that change. Because we are not given the sequence in which the groups will leave. So that also needs to be considered.
One more note is that in question they are given the lift will not stop at $2^{\text{nd}}$ floor that’s why only that floor is excluded. But if the words are like upto $2^{\text{nd}}$ floor then we will exclude 2 floors $1^{\text{st}}$ and $2^{\text{nd}}$.
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