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Chapter 5: Problem 443

In a circus there are 10 cages for accommodating 10 animals out of these 4 cages are so small that five out of ten animals can not enter into them. In how many ways will it be possible to accommodate 10 animals in these 10 cages? (a) 66400 (b) 86400 (c) 96400 (d) 46900

Short Answer

Expert verified
In summary, by placing 5 animals in the 4 small cages, we have 120 arrangements. Then, by putting the remaining 6 animals in the 6 large cages, we have 720 arrangements. By multiplying these two numbers, we find a total of 86,400 different ways to place the animals in the cages. The answer is (b) 86400.

Step by step solution

01

Place animals in the small cages

We have 4 small cages and 5 animals that can fit in them. To place these 5 animals in the 4 cages, we can use the formula for the number of arrangements: n!/(n-r)!, where n is the total number of animals, and r is the number of cages. In our case, n = 5 (animals that can fit into small cages) and r = 4 (number of small cages). So we have: \( Arrangements_1 = \frac{5!}{(5-4)!} = \frac{120}{1} = 120 \)
02

Place animals in the large cages

Now we have 6 animals left (10 - 4 who are already in small cages), and 6 large cages. All of these animals can fit into these large cages. To place these 6 animals in the 6 cages, we can use the same formula for the number of arrangements. In this case, n = 6 (animals left) and r = 6 (number of large cages). So we have: \( Arrangements_2 = \frac{6!}{(6-6)!} = \frac{720}{1} = 720 \)
03

Multiply the arrangements to get total permutations

We just need to multiply the arrangements of the small cages and the arrangements of the large cages to get the total number of permutations: \( TotalPermutations = Arrangements_1 \times Arrangements_2 = 120 \times 720 = 86400 \) The answer is (b) 86400.

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Most popular questions from this chapter

if \(a_{n}={ }^{n} \sum_{r=0}\left[1 /\left({ }^{n} C_{r}\right)\right]\) then value of \(^{n} \sum_{r=0}\left[(n-2 r) /\left({ }^{n} C_{r}\right)\right]\) is (a) \((\mathrm{n} / 2) \mathrm{a}_{\mathrm{n}}\) (b) \((1 / 4) \mathrm{a}_{\mathrm{n}}\) (c) \(\mathrm{n} \cdot \mathrm{a}_{\mathrm{n}}\) (d) none of these.

A car will hold 2 in the front seat and 1 in the rear seat. If among 6 persons 2 can drive, than number of ways in which the car can be filled is (a) 10 (b) 20 (c) 30 (d) 40

The number of numbers greater than 3000 , which can be formed by using the digits \(0,1,2,3,4,5\) without repetition is (a) 1240 (b) 1280 (c) 1320 (d) 1380

The number of integer \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\), such that \(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=20\) and \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \geq 0\) is (a) \({ }^{24} \mathrm{C}_{3}\) (b) \({ }^{25} \mathrm{C}_{3}\) (c) \({ }^{26} \mathrm{C}_{3}\) (d) \({ }^{27} \mathrm{C}_{3}\)

If the different permutations of all the letters of the word EXAMINATION are listed in a dictionary then how many words are there in this list before the first word begins with \(\mathrm{E}\) ? (a) 907200 (b) 970200 (c) 922700 (d) 709002
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