The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is
A.
$$9!\, \times 10!$$
B.
$$5{\left( {9!} \right)^2}$$
C.
$${\left( {9!} \right)^2}$$
D.
None of these
Answer :
$$5{\left( {9!} \right)^2}$$
Solution :
Ten pearls of one colour can be arranged in $$\frac{1}{2} \cdot \left( {10 - 1} \right)!$$ ways. The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour $$= 10!.$$
∴ the required number of ways $$ = \frac{1}{2} \times 9!\, \times 10!.$$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is