Question

The letters of the word “FLOWER” are taken 4 at a time and arranged in all possible ways. The number of arrangements which begin with ‘F’ and end with ‘R’ is
A. 20
B. 18
C. 12
D. 14

Answer 1

Hint: Total no. of letters in “FLOWER” is 6. We need to select four letter words in which F is the starting letter and R is the ending letter. This means out of 6 letters 2 (F, R) are already used. So we have 4 remaining. First and last letters of the 4-letter words are fixed and the remaining 2 letters must be selected from the remaining 4 letters. Use combinations to select the letters.

Complete step by step solution:
We are given that 4 letters of the word “FLOWER” are taken at a time.
We need to find the no. of arrangements of the 4-lettered words which begin with F and end with R.
Four letter words must be in the form F_ _ R.
So first and last words are already selected, we have to select the remaining 2 from the remaining 4 letters of the word, L, O, W, E.
The 2nd letter can be selected from 4 letters and the 3rd letter can be selected from 3 letters.
So total no. of arrangements of the 4-letter word that begin with F and end with R is $1 \times {}_{}^4C_1^{} \times {}_{}^3C_1^{} \times 1$
${}_{}^nC_r^{} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}},{}_{}^4C_1^{} = 4,{}_{}^3C_1^{} = 3$
$ \Rightarrow 1 \times 4 \times 3 \times 1 = 12\;ways$
So, the correct answer is “Option C”.

Note: A Permutation is arranging the objects in order. Combinations are the way of selecting the objects from a group of objects or collection. When the order of the objects does not matter then it should be considered as Combination and when the order matters then it should be considered as Permutation. Do not confuse using a combination, when required, instead of a permutation and vice-versa