I. The number of ways in which $$4$$ letters can be posted in $$5$$ letter boxes in $${4^5}$$ ways.II. If words are formed by taking only $$4$$ at a time out of the letters of the word PHYSICS, then the number of words in which Y occurs, is $$288$$.Which of the above statements is correct $$?$$A) Only IB) Only IIC) Both I and IID) Neither I nor II

Question

I. The number of ways in which $$4$$ letters can be posted in $$5$$ letter boxes in       $${4^5}$$ ways.II. If words are formed by taking only $$4$$ at a time out of the letters of the word PHYSICS, then the number of words in which Y occurs, is $$288$$.Which of the above statements is correct $$?$$A) Only IB) Only IIC) Both I and IID) Neither I nor II

Vedantu Content Team · Accepted Answer

Hint: First we know a combination is a mathematical process that determines the number of possible arrangements in a collection of items where we select the items in any specific order. The combination formula $${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\;\;r!\;}}$$ where, $$n$$ is the total items in the set and $$r$$ is the number of items taken for the permutation to find the value of $$n$$.Each of the letters can be posted in any of the letter boxes. This means that every letter can be posted in a number of letter boxes. If $$n$$ letters can be posted in $$m$$ letter boxes, then the number of ways of posting  $$n$$ letters in $$m$$ boxes is $${m^n}$$. In second part the number of words formed by taking only $$4$$ where Y occurs at a time out of the letters of the word PHYSICS is equal to sum of the number of words formed with $$3$$ different letters along with Y and the number of words with two same letters along with Y.Complete step-by-step answer:I.Each of the letters can be posted in any of the $$5$$ letter boxes. This means that every letter can be posted in $$5$$ ways.$$\therefore $$Total number of ways of posting $$4$$ letters in $$5$$ letter boxes$$ = 5 \times 5 \times 5 \times 5$$$$ = {5^4}$$. II.The word ‘PHYSICS‘  contains $$7$$ letters in which S is repeated two times.We have to always take the Y letter out of $$4$$ at a time out of the letters of the word PHYSICS, then the number of words in which Y occurs is equal to the sum of the all different characters along with Y and Two are the same along with Y. If we take 3 different letters with Y (except S letter), then the total number of words will be $$ = {}^5{C_3} \times 4! = 5 \times 2 \times 4! = 240$$.If we take two same letters S with y, then then total number of words will be (In each word S letter repeated twice) $$ = {}^4{C_1} \times \dfrac{{4!}}{{2!}} = 4 \times 4 \times 3 = 48$$. $$\therefore $$The number of words in which Y occurs$$ = 240 + 48 = 288$$.So, the correct answer is “Option B”.Note: Note that each of the different arrangements which can be made by taking some or all of a number of things is called a permutation. The number of permutations of $$m$$ different things taken $$n$$ at a time, allowing repetitions is $${m^n}$$. The number of permutations of $$n$$ different things taken all at a time is $${}^n{P_n} = n!$$.