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How many different permutations can be made with the letters of the word 'CONSTANT' so that vowels are always together?
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in this problem, we are given the word constant that has warbles as zero and A. And consonants as see and yes, D. And entity Yeah. eight letters in total. Now we want the Vogels to always be together. So we have 123456 I'm making only seven spaces because the mobile, which is E. Always comes together as one entity. So we have one entity, oh and six different entities of consonants. So we have seven spaces that we have to fill. So the number of ways that we can fill it is seven factorial, but since oh, a can be interchanged as A. A. Also. So we multiply it by two. No en entity. The both occurred twice. So we have to divide by two factorial twice to account for the reputation of N and D. So these are the total um rubies. It's calculated this this is nothing but seven factorial by two, which is equal to now. seven factorial is nothing with 5040. Not divided by two ways 2 5- zero. This comes as 25 does it not? So
How Many Strings Can Be Formed With The Given Word :
Here we are going to see some practice questions based on the concept fundamental principle of counting.
How many strings can be formed with the given word - Questions
Question 1 :
How many strings can be formed using the letters of the word LOTUS if the word [i] either starts with L or ends with S? [ii] neither starts with L nor ends with S?
Solution :
[i] either starts with L or ends with S?
Case 1 :
Number of words starts with L,
L ___ ___ ___ ___
1st 2nd 3rd 4th 5th
1st dash --> We have 1 option [L]
2nd dash --> We have 4 options
3rd dash --> We have 3 options
4rd dash --> We have 2 options
5rd dash --> We have 1 option
= 1 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1
= 24 words
Case 2 :
Number of words ends with S,
___ ___ ___ ___ S
1st 2nd 3rd 4th 5th
5rd dash --> We have 1 option [S]
1st dash --> We have 4 options
2nd dash --> We have 3 options
3rd dash --> We have 2 options
4rd dash --> We have 1 options
= 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 1
= 24 words
Case 3 :
Number of words starts with L and ends with S.
L ⋅ 3 ⋅ 2 ⋅ 1 ⋅ S
= 6
Either starts with L or ends with S = 24 + 24 - 6
= 42 words
[ii] neither starts with L nor ends with S?
Total number of words formed without restriction = 5!
= 120 words
Number of words starts with L nor S
= Total number of words - Number of words either starts with L or ends with S
= 120 - 42
= 78 words
Question 3 :
[i] Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
[ii] In how many ways 10 pigeons can be placed in 3 different pigeon holes ?
[iii] Find the number of ways of distributing 12 distinct prizes to 10 students?
Solution :
[i] Each question is having 4 options.There are 4 ways to answer each question.
1st question = 4 ways
2nd question = 4 ways .........
Total number of ways = 46
[ii] In how many ways 10 pigeons can be placed in 3 different pigeon holes ?
Note : If n different objects are to be placed in m places, then the number of ways of placing is mn.
10 - Number of pigeons [different things]
3 - Number of places [pigeon holes]
Applying the formula mn,
Total number of ways = 310
[iii] 12 - Number of prizes [different things]
10 - Number of place [students]
Applying the formula mn,
Total number of ways = 1012
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