When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
∴ Required number of ways = (120 x 6) = 720.
∴ Required number of ways  = (^{3}C_{1} x ^{6}C_{2}) + (^{3}C_{2} x ^{6}C_{1}) + (^{3}C_{3})  


= (45 + 18 + 1)  
= 64. 
= (^{7}C_{3} x ^{4}C_{2})  


= 210. 
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves 
= 5! 
= 5 x 4 x 3 x 2 x 1  
= 120. 
∴ Required number of ways = (210 x 120) = 25200.
∴ Required number of ways =  6!  = 360. 
(1!)(2!)(1!)(1!)(1!) 
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = ^{3}P_{3} = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = ^{3}P_{3} = 3! = 6.
Total number of ways = (6 x 6) = 36.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
∴ Required number of ways = (120 x 6) = 720.
Required number of ways  = (^{8}C_{5} x ^{10}C_{6})  
= (^{8}C_{3} x ^{10}C_{4})  


= 11760. 
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =  7!  = 2520. 
2! 
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in  5!  = 20 ways. 
3! 
∴ Required number of ways = (2520 x 20) = 50400.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
∴ Required number of numbers = (1 x 5 x 4) = 20.
Required number of ways = (^{7}C_{5} x ^{3}C_{2}) = (^{7}C_{2} x ^{3}C_{1}) =  ❨  7 x 6  x 3  ❩  = 63. 
2 x 1 
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