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Using the digits 0,1,2,...9 how many 7-digit numbers can be formed if the first digit cannot be 0 or 9 and if the last digit is greater than or equal to 2 and less than or equal to 3? Repeated digits are allowed. Solution Let the number be 'abc'. Then, $$2<a\times\ b\times\ c<7$$. The product can be 3,4,5,6. We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria. We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers. We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers. Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.
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Hint:let the number be $\overline{xyz}$ Now if $x\cdot y\cdot z=p$ where $p=3,5$ then we $x,y,z$ must be a permutation of $1,1,p$ Hence number of such numbers is $2\cdot \frac{3!}{2!}$ Now if $x\cdot y\cdot z=4,6$ then $x,y,z$ must be a permutation of either $1,1,4$ or $2,2,1$ or $1,1,6$ or $2,3,1$ .Each of which the number of cases can be easily found......
Answer (Detailed Solution Below) 21
12 Questions 36 Marks 20 Mins
Calculation: Let the 3-digit number be abc, ⇒ a × b × c = 3 or 4 or 5 or 6
⇒ Total numbers = 3 + 6 + 3 + 9 = 21 ∴ There are 21 digits whose products of their digits is more than 2 but less than 7. India’s #1 Learning Platform Start Complete Exam Preparation
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How many 3 digit numbers are there for which the product of their digits is more than 2 but less?
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