nCr and nPr 1

(Q1)
The arrangement:
3 people (D, F, G) of 151 cm
2 people (C, E) of 153 cm
A of 158 cm
B of 159 cm

No. of permutations of the 3 151-cm people = 3P3 = 3! = 6
No. of permutations of the 2 153 cm people = 2P2 = 2! = 2

Total no. of permuations = 6 x 2 = 12


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(Q2)
(a)
To put one Chinese book as the first book,
no. of permutations = 4P1 = 4!/(4 - 1)! = 4

To put the rest 8 books in order,
no. of permutations = 8P8 = 8! = 40320

If the first book is a Chinese book, no. of permutations
= 4 x 40320
= 161280

(b)
To put 2 Chinese books as the first 2 books,
no. of permutations = 4P2 = 4!/(4 - 2)! = 12

To put the rest 7 books in order,
no. of permutations = 7P7 = 7! = 5040

If the first 2 books are Chinese books, no. of permutations
= 12 x 5040
= 60480

(c)
To put 3 Chinese books as the first 3 books,
no. of permutations = 4P3 = 4!/(4 - 3)! = 24

To put the rest 6 books in order,
no. of permutations = 6P6 = 6! = 720

If the first 3 books are Chinese books, no. of permutations
= 24 x 720
= 17280


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(Q3)
Among the 7 numbers, there are 3 odd numbers and 4 even numbers.

To select an odd number to be the units digit,
no. of ways = 3P1 = 3!/(3 - 1)! = 3

To select 2 numbers from the rest 6 numbers as the tens digit and the hundreds digit,
no. of ways = 6P2 = 6!/(6 - 2)! = 30

Number of 3-digit odd numbers can be formed
= 3 x 30
= 90


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(Q4)
Let n be the number of cakes available.

No. of permutation when 2 different kinds of cakes are chosen:
= nP2
= n!/(n - 2)!
= n(n - 1)

No. of permutation when 3 different kinds of cake are chosen:
= nP3
= n!/(n - 3)!
= n(n - 1)(n - 2)

n(n - 1) + n(n - 1)(n - 2) = 576
n(n - 1)[1 + (n - 2)] = 576
n(n - 1)² = 9 x 8²
n = 9

Number of different kinds of cakes available = 9