
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 find p:q.
A. 6:5
B. 3:2
C. 4:3
D. 5:3
Answer
162.6k+ views
Hint: First we find the least number that is made by using the given number. Then find the total number of numbers that are made by a given number using the factorial formula. Then we will the number of numbers that lie between 30000 and 90000 by finding the number of numbers that started with 3,5, and 7. Then using the value of p and q we will find p:q.
Formula Used: The number of arrangements of n objects is n!.
Complete step by step solution: Given numbers are: 2, 3, 5, 7, 9.
The least number that is made by the given number is 23579.
The number of numbers that are made by given numbers is \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\].
Therefore the value of p is 120.
The numbers that lie between 30000 and 90000 must start with 3,5,7. If the number starts with 2 then it will be less than 30000. If the number starts with 9 then it will be greater than 90000.
When the digit at ten thousand is 3:
The rest 4 digits we can choose from 2, 5, 7, 9.
The number of numbers is \[4! = 4 \times 3 \times 2 \times 1 = 24\]
When the digit at ten thousand is 5:
For the rest 4 digits we can choose from 2, 3, 7, 9.
The number of numbers is \[4! = 4 \times 3 \times 2 \times 1 = 24\]
When the digit at ten thousand is 7:
For the rest 4 digits, we can choose from 2,3, 5, 9.
The number of numbers is \[4! = 4 \times 3 \times 2 \times 1 = 24\]
The total number of numbers that lie between 30000 and 90000 is 24+24+24=72.
Therefore the value of q is 72.
The ratio of p to q is 120:72 = 5:3
Option ‘D’ is correct
Note: Students often confused with combination and arrangement. If we arrange r objects from n objects then we apply a combination. If we need to use all objects to make arrangements we apply the factorial formula. In the given question we need to apply the factorial formula.
Formula Used: The number of arrangements of n objects is n!.
Complete step by step solution: Given numbers are: 2, 3, 5, 7, 9.
The least number that is made by the given number is 23579.
The number of numbers that are made by given numbers is \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\].
Therefore the value of p is 120.
The numbers that lie between 30000 and 90000 must start with 3,5,7. If the number starts with 2 then it will be less than 30000. If the number starts with 9 then it will be greater than 90000.
When the digit at ten thousand is 3:
The rest 4 digits we can choose from 2, 5, 7, 9.
The number of numbers is \[4! = 4 \times 3 \times 2 \times 1 = 24\]
When the digit at ten thousand is 5:
For the rest 4 digits we can choose from 2, 3, 7, 9.
The number of numbers is \[4! = 4 \times 3 \times 2 \times 1 = 24\]
When the digit at ten thousand is 7:
For the rest 4 digits, we can choose from 2,3, 5, 9.
The number of numbers is \[4! = 4 \times 3 \times 2 \times 1 = 24\]
The total number of numbers that lie between 30000 and 90000 is 24+24+24=72.
Therefore the value of q is 72.
The ratio of p to q is 120:72 = 5:3
Option ‘D’ is correct
Note: Students often confused with combination and arrangement. If we arrange r objects from n objects then we apply a combination. If we need to use all objects to make arrangements we apply the factorial formula. In the given question we need to apply the factorial formula.
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