In How many ways can 7 people park cars in slots allotted to them?

Why do we use the combination formula instead of the permutation formula?

Hi Zynxu,
Question asks:

In how many ways can they park their cars so that exactly five persons park their cars in the slots allotted to them?

Or in other words, we can say in how many ways two parks will be parked in the wrong slot?

For example. car no. 6 and car no. 7 are parked in the wrong slot. This we won't count as 2, rather we will count as 1 way only.
Hence you need to apply combination.

Another way we can think as follows:

Suppose car no. 1 in in the wrong place, so all possible combination for the car no. 1 will be:

(1,2), (1,3), (1,4), (1,5), (1,6), (1,7) --- Total 6 ways.

For car no. 2, the following possibilities are there:

(2,1), (2,3), (2,4), (2,5) (2,6), (2,7) --- Total 6 ways.

Similarly, for all other cars, we'll have 6 such possibilities.

Hence the total number of ways = 7*6 = 42 ways.

But, we are double counting each option. Hence, we need to divide the final answer by 2.

The required number of ways = 42/2 = 21 ways.

Hope this helps.

Thanks.

Pick any 5 person from 7 person = 5C7 = 7*6/3= 21 ways
These people parked on their respective slots.

Remaining 2 people & 2 Slots. only one way to park (i.e on other's parking space) possible. =1 way
Total possible ways= 21*1=21

Ans.B

Can you provide more information on your logic/intent behind that calculation?

7!/(7-5)!5!
This will give 21

Posted from my mobile device

Super. I did the 7C5 but then i multiplied it by 2 thinking that the remaining cars can be parked in two ways whereas there is only one arrangement for them, as the last one would mean that they are in their correct places, which does not fit the answer. (a very confusing statement!!!)

shouldn't it be 7C5????? and 42/2.... not 42/3......


☺☺

Posted from my mobile device

What I simply did was added the pairs of remaining 2 people who will have to park in each others spots:
Such as: 12,13,14,15,16,17,
23,24,25,26,27,
34,35,36,27,
45,46,47,
56,57,
67.

Hence option B. 21.
Notice that 12 is the same as 21, 23 is the same as 32, and so on. Hence, those need to be ignored.
Hope this helps

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