Concept:
- The number of ways in which r objects can be arranged in n places is nPr = \(\rm \dfrac{n!}{(n - r)!}\).
- The number of ways in which n objects, out of which p, q, r etc. are of same type, can be arranged is: \(\rm \dfrac{n!}{p!\ q!\ r!\ ...}\).
- The number of ways in which r distinct objects can be selected from a group of n distinct objects is nCr = \(\rm \dfrac{n!}{r!(n - r)!}\).
- n! = 1 × 2 × 3 × ... × n.
- 0! = 1.
- If some objects cannot be together, then first arrange the remaining objects, and then arrange the objects which cannot be together in the empty spaces between them.
Calculation:
The word PROBABILITY has 11 letters out of which 4 (O, A, I, I) are vowels and 7 (P, R, B, B, L, T, Y) are consonants.
Since no two vowels can be together, we will first arrange the remaining 7 consonants.
The consonant can be arranged in \(\rm \dfrac{7!}{2!}=2520\) ways.
Now, the 4 vowels can be arranged in any of the 8 spaces between the consonants, as illustrated by a _ below:
_ P _ R _ B _ B _ L _ T _ Y _.
Number of ways of selecting some 4 places out of the 8 places = \(\rm ^8C_4 = \dfrac{8!}{4!(8 - 4)!}=70\).
Number of ways in which the 4 vowels can be arranged in these 4 places = \(\rm \dfrac{4!}{2!}=12\)
∴ The total number of possible arrangements = 2520 × 70 × 12 = 21,16,800.
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Solution
The correct option is C
14400
The explanation for the correct answer.
Find the number of ways such that no two vowels occur together.
Given: TRIANGLE
The total number of letters in the word triangle is 8.
The total number of words that can be formed =8!=40320.
The total number of words in which two vowels occur together =C2×7!×2!= 302403.
The number of ways where all vowels occur together =C3×6!×3!= 43203.
Therefore required number of words =40320-30240+4320=14400.
Hence option (C) is the correct answer.
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Christopher G.
Algebra
6 months ago
We don’t have your requested question, but here is a suggested video that might help.
In how many ways can the letters of the word number be arranged?
FRACTION
vowels = AIO, let k
FRCTNK
6! => 6X5X4X3X2X1= 720
so, 720 ways can be arranged without two vowels are together.
total ways in which the letters of the word FRACTION can be arranged = 8! = 40320 vowels : A I O ( let us consider them as 1 letter instead of 3 letters and let this new letter be * ) consonants : F R C T N now the new word thus formed will be : * F R C T N the letters of this new word can be arranged in 6! ways. not only that * can arrange itself in 3! ways so the total ways in which all the vowels are together = 6! x 3! = 4320
so total words that can be formed so that no two vowels are together = 40320 - 4320 = 36000
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Ravi Ranjan Ojha