5 - digit numbers are to be formed using 2, 3, 5, 7, 9 without repeating the digits. If $$p$$ be the number of such numbers that exceed 20000 and $$q$$ be the number of those that lie between 30000 and 90000, then $$p : q$$ is :
A.
$$6 : 5$$
B.
$$3 : 2$$
C.
$$4 : 3$$
D.
$$5 : 3$$
Answer :
$$5 : 3$$
Solution :
\[p:\begin{array}{*{20}{c}}
{TTH}&{TH}&H&T&0&{{\rm{place}}}\\
5&4&3&2&1&{{\rm{ways}}}
\end{array}\]
Total no. of ways $$= 5! = 120$$
Since all numbers are $$> 20,000$$
$$\therefore $$ all numbers 2, 3, 5, 7, 9 can come at first place.
\[q:\begin{array}{*{20}{c}}
{TTH}&{TH}&H&T&0&{{\rm{place}}}\\
3&4&3&2&1&{{\rm{ways}}}
\end{array}\]
Total no. of ways $$= 3 × 4! = 72$$
($$\because $$ 2 and 9 can not be put at first place)
So, $$p : q = 120 : 72 = 5 : 3$$
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