skysailor
Could someone help me understand how the total number of cases is 5!/2!? I don't understand the reasoning behind that.
Hi
skysailor Thanks for your query.
To get a better understanding of the “why” behind the formula used, check out the example given below.
EXAMPLE: How many 4-letter words can be formed using the letters of the word POOR? One basic concept on permutations tells us that if we need to arrange 4 distinct elements, we can do so in
4! (= 24) ways. So, if all 4 letters were distinct, there would be 24 possible words that can be formed using those letters.
But, as you must have observed, in the word POOR, we have 2 identical letters. You must first understand that if we still do 4!, it will involve a lot of extra counting – that is it will count the same word more than once. See how:
- Suppose that in the word POOR, we keep P and R fixed and just change the positions of the two Os. Then, will we get a new word? NO!
- POOR is the same as POOR.
- Similarly, let’s consider another arrangement of the letters, P, O, O, R, say, ROOP. Even now, ROOP is the same as ROOP.
Since ‘
O’ is not different from ‘
O’, it doesn’t make sense to count P
OOR and P
OOR as two different words, or R
OOP and R
OOP as two different words.
But when we calculate 4!, it does count these words as distinct. And hence, we need to
factor these out.
Factor out repetitions: Since for every fixed position of P and R, the
two Os can be arranged in 2! ways, we will divide the total of 4! by 2!.
Note that had there been three Os, we would have divided it by 3! to factor out all arrangements obtained by ONLY arranging the Os. (Again, because these three Os can be arranged in 3! ways, for every fixed position of the other elements.)
To summarize the example, we can say the total number of words that can be formed using the letters of the word POOR = \(\frac{4! }{ 2!}\) = 12.
GENERAL PRINCIPLE RESULT: If we must find the total number of arrangements of
‘n’ items, out of which
- ‘p’ items are identical, then the total possible arrangements = n! / p!
- ‘p’ items are identical to each other, and ‘q’ items are identical to each other, then the total possible arrangements = n!/(p!q!), and so on.
APPLICATION ON YOUR QUESTION: Finally, let’s apply this to our question at hand.
We need to find the total number of possible 5-letter words that can be formed using the letters of the word
DIGIT. So,
- Total letters available (n) = 5
- Repetitive letters count (p) = 2 (two Is)
- Hence, the total number of words = 5! / 2! = 5 × 4 × 3 = 60.
Hope this helps!
Best,
Aditi Gupta
Quant expert,
e-GMAT