How many different letter arrangements can be formed from the letters pineapple

Note: A permutation is an act of arranging the objects or numbers in order while Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.

The formula for permutations is given by: \[{}^n{P_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{\left[ {n - r} \right]!}}\]
The formula for combinations is given by: \[{}^n{C_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{r!\left[ {n - r} \right]!}}\]
In the case of repetitions while arranging:
Since in the word BANANA, there are 3 A’s and 2 N’s, therefore while calculating the number of arrangements one must consider the repetitions of the letters. Note that if there are n things to be arranged in a row, among which things are of one kind, b things are of another kind, and c things are of another, then the total number of arrangements is given by $\dfrac{{n!}}{{a!b!c!}}$.

The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares.

There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$

• The number of permutations of n things taken all together, when p of the things are alike of one kind, q of them alike of another kind, r of them alike of a third kind and the remaining all different is n! / [p! q! r!]
• The number of permutations of n objects, of which m are of one kind and the rest n -m of another kind, taken all at a time is n! / [m! [n-m]!]

Illustrative Examples

Example

In how many ways can the letters of the word permutations be arranged such that

1. there is no restriction
2. P comes before S
3. words start with P and end with S
4. T's are together
5. vowels are together
6. order of vowels remains unchanged?

Solution

The given word has 12 letters -two Ts and 10 different letters.

1. Total number of arrangements is12!/2! = 6. 11!
2. Out of above, P comes before S in half the arrangements. Hence required number of arrangements = 3. 11!
3. As position of P and S is fixed, remaining 10 letters [that is, two T's and eight other different letters] can be arranged in 10!/2! = 5. 9! ways.
4. Considering two T's as a block, we have to arrange 11 different things, which can be done in 11! ways.
5. Considering the five vowels in given letter -E, U, A, I, O as a block, we have 8 things having 2 alike things [T's]. So this can be arranged in 8!/2! = 4. 7! ways. Now within the block, 5 different vowels can be arranged in 5! ways. Hence required number of arrangements is 4. 7!5! ways.
6. If order of five vowels has to remain unchanged, we can consider them like five alike things, so only one ordering is possible. Thus we have 12 things of which 2 are alike and 5 are alike. Hence required number of arrangements is 12!/[2! 5!]

Exercise

1. How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used?
2. In how many ways can five red marbles, two white marbles and three blue balls be arranged in a row?
3. In how many ways can you arrange six identical coins in a row so that you get exactly 2 heads?
4. Find the number of arrangements that can be made out of the following words:
   [i] BANANA
   [ii] APPLE
   [iii] PINEAPPLE
   [iv] INDEPENDENCE
   [v] ASSASSINATION
5. How many arrangements can be made with the letters of the word MATHEMATICS if
   [i] there is no restriction
   [ii] vowels occur together
   [iii] all vowels don't occur together
   [iv] consonants occur together
   [v] M is at both extremes
   [vi] order of vowels remains unchanged?
6. How many different numbers can be formed out of all the digits of 111223? How many of these are greater than 300000?
7. How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?
8. How many numbers greater than 100000 can be formed by using the digits 2, 4, 2, 3, 0, 2 taken all together?
9. How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places?
10. If permutations of all the letters of the word AGAIN are arranged in dictionary order, which is the fiftieth word?
11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's occur together?

Answers

1. 1260                       2. 2520            3. 15
4. [i] 60                      [ii] 60                [iii] 30240        [iv] 1663200        [v] 10810800
5. [i] 4989600            [ii] 120960        [iii] 4868640     [iv] 75600
    [v] 90720               [vi] 415800        6. 60, 10          7. 360
8. 100                        9. 18                10. NAAIG      11. 151200

How many four letters word can be formed out of the word pineapples?

How many letter arrangements can be made from letters?

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