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Permutation Of Objects When Few Are Identical - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
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93 Questions around this concept.
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A seven digit number is formed using digits 3,3,4,4,4,5,5. The probability that number so formed is divisible by 2, is:
Consider the word "ENGINEERING". In how many ways can the letters be arranged such that all the vowels always come together, but the two 'E's are not adjacent and the three 'N's are in alphabetical order?
How many different words can be formed with the letters of the word "ENCAPSULATE" such that: (i) Four vowels occupy the odd places? (ii) The word begins with the letter N? (iii) The word begins with the letter E and ends with the letter T?
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In how many ways can the letters of the word "APPRECIATE" be arranged such that the word should have the first letter P?
How many ways can the letters of the word "ARRANGE" be arranged such that the two R's are never together?
In how many ways can the letters of the word "PERMUTE" be arranged if all the letters must be used?
Find the number of ways of arranging the letters of the word!
"MISSISSIPPI"
What is the no of ways in which 7 objects can be divided into 2 groups of 3 & 4 objects each?
Several ways in which 3 boys and 3 girls (all of different heights) can be arranged in a line so that boys as well as girls, among themselves, are in decreasing order of height (from left to right) is:
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Try NowNumber of arrangements of digits 1,1,0,0,0,4,4,4 that start and end with 4, are
Concepts Covered - 1
PERMUTATION OF OBJECTS WHEN FEW ARE IDENTICAL
If there n objects of which p objects are of one type, q objects of another type, r objects of yet another type, and all others are distinct, the total number of ways of arranging all the objects is $\frac{\mathrm{n}!}{\mathrm{p}!\mathrm{q}!\mathrm{r}!}$
Proof:
Suppose the total number of permutations is $x$, now if we replace all $p$ identical objects by $p$ different objects then we have $x \times$ p arrangements. The number of arrangements, if we do the same thing with $q$ and $r$, will be $x \times p!\times q!\times r!$
Now, we have replaced all identical objects and we are left with $n$ different object which can be arranged in n! Ways.
Hence, $x \times p!\times q!\times r!=n!$
So, $\mathrm{x}=\frac{\mathrm{n}!}{\mathrm{p}!\mathrm{q}!\mathrm{r}!}$
Example: In how many ways can the letters of the word “MISSISSIPPI” be arranged?
Solution: repeated letters I = 4 times, S = 4 times and P=2 times
So using the above formula we have $x=\frac{11!}{4!\times 4!\times 2!}$
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