1.Find the number of ways in which the letters of the word (I) honesty (II) obedience can be arranged in a row taken all at a time. 2. How many numbers greater than 67000 can be formed by using the digits 5,6,7,8 and 9 ,no digit is repeated in any number?3. How many odd numbers are there with three digits such that if 4 is one of the digits then 6 is the next digit?4. How many words can be formed from the letters of word 'MALENKOV' if no two vowels occur together? Please give the full explanations for all the above questions.
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1)(I)honesty=honesty consist 7 letters and no letter is repeated hence= 7!=5040
(II) obedience=it consist of 9 letters with e letter is repeated 3 times hence=9!/3!=9×8×7×6×5×4=60480
2)no. which is greater than 67000 can be form by keeping 4 no. in 1st place i.e.6 is the 1st place digit by replacing 6 by 6,7,8,9 means 4×replacing 2nd place i.e.7 by 7,8,9 and remaining can be replace by any 5 no.,hence ans is 4×3×5×5×5="1500" since we can repeat the no.
3) let's 4 is in 100nd place i.e 4** and 6 is in 10th place means 46* then there are 5 odd no. and if we consider like this 64* then there are also 5 odd no. hence total odd no.s are "10",we can not consider 4 and 6 in unit place then it becomes even no.
4)MALENKOV=no. of vowels are 3 A,E,O.Our main aim is they shouldn't come together.
3 vowels can be place alternatively at 3 different places and remaining are place as follows
3×5×3×5×3×5×3×5=50625
3-vowels
5-remaining letters m,l,n,k,v
If you find my ans helpful please mark me as "brainlist"
(II) obedience=it consist of 9 letters with e letter is repeated 3 times hence=9!/3!=9×8×7×6×5×4=60480
2)no. which is greater than 67000 can be form by keeping 4 no. in 1st place i.e.6 is the 1st place digit by replacing 6 by 6,7,8,9 means 4×replacing 2nd place i.e.7 by 7,8,9 and remaining can be replace by any 5 no.,hence ans is 4×3×5×5×5="1500" since we can repeat the no.
3) let's 4 is in 100nd place i.e 4** and 6 is in 10th place means 46* then there are 5 odd no. and if we consider like this 64* then there are also 5 odd no. hence total odd no.s are "10",we can not consider 4 and 6 in unit place then it becomes even no.
4)MALENKOV=no. of vowels are 3 A,E,O.Our main aim is they shouldn't come together.
3 vowels can be place alternatively at 3 different places and remaining are place as follows
3×5×3×5×3×5×3×5=50625
3-vowels
5-remaining letters m,l,n,k,v
If you find my ans helpful please mark me as "brainlist"
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