Question

The number of combinations that can be made by taking 4 letters of the word ‘COMBINATION’ is
A) 70
B) 63
C) 3
D) 136

Answer 1

We will first divide our solution in three cases: both like letters twice, one pair of like and other alike and all alike. We will then individually find a number of ways in each case and then sum them up to get the required answer.

Complete step-by-step answer:
We have the word “COMBINATION”. Here, we have 1 C, 2 O’s, 1 M, 1 B, 2 I’s, 2 N’s, 1 A and 1 T.
We basically have 11 letters among which 8 are distinct.
Now, we will have 3 cases to choose letter:
Case 1: We can select 2 pair of 2 letters which are occurring twice in “COMBINATION”.
We have 3 such pairs that are O’s, I’s and N’s.
We need to select 2 from these 3 to get 4 letters.
Hence, the number of ways we can do this is $^3{C_2}$.
We know that $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ and $n! = n(n - 1)(n - 2)......1$. ……(A)
So, $^3{C_2} = \dfrac{{3!}}{{2!(1)!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1}} = 3$.
Hence, case 1 arises in 3 possible ways. ………….(1)
Case 2: We can select 1 pair of 2 letters which are occurring twice in “COMBINATION” and 2 distinct letters.
We have 3 such pairs that are O’s, I’s and N’s.
We need to select 1 from these 3 to get 4 letters.
Hence, the number of ways we can do this is $^3{C_1}$.
Using (A), we will get:-
So, $^3{C_1} = \dfrac{{3!}}{{1!(2)!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1}} = 3$.
We also need to select 2 distinct letters from 1 C, 1 M, 1 B, 1 A and 1 T and 2 pair of like letters.
This means we need to select 2 letters from 7 letters.
Hence, the number of ways we can do this is $^7{C_2}$.
Using (A), we will get:-
So, $^7{C_2} = \dfrac{{7!}}{{2!(5)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1 \times 2 \times 1}} = 21$.
Hence, case 2 gives rise to $3 \times 21 = 63$ possible ways. ………….(2)
Case 3: We can 4 distinct letters which are occurring in “COMBINATION”.
We also need to select 4 distinct letters from 1 C, 1 M, 1 B, 1 A and 1 T and 3 pair of like letters.
This means we need to select 4 letters from 8 letters.
Hence, the number of ways we can do this is $^8{C_4}$.
Using (A), we will get:-
So, $^8{C_4} = \dfrac{{8!}}{{4!(4)!}} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 4 \times 3 \times 2 \times 1}} = 70$.
Hence, case 3 gives rise to 70 possible ways. ………….(3)
Combing (1), (2) and (3), we will get:-
Total combination = 3 + 63 + 70 = 136.

So, the correct answer is “Option B”.

Note: The students must note that when in case 2, we had 9 letters to choose, but we took 7 instead of 9, we did that to eliminate the possibility of repetition because we considered that in case 1 and the same thing is done by us in case 3 as well.
The students must note that “permutation and combination” makes our tasks so easy. We cannot go on writing all the words and combinations possible out of this word. So, we have an easy way to find its number at least.