How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
 Number of places [(x), (y) and (z)] for them = 3Repetition is allowed and the 3-digit numbers formed are oddNumber of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd) m = 3 Number of ways of filling box (y) = 6 (∴ Repetition is allowed)n = 6 Number of ways of filling box (z) = 6 (∵ Repetition is allowed)p = 6 ∴ Total number of 3-digit odd numbers formed = m x n x p = 3 x 6 x 6 = 108(b) Number of ways of filling box (x) = 3 (only odd numbers are to be in this box )m = 3 Number of ways of filling box (y) = 5 (∵ Repetition is not allowed)n = 5 Number of ways of filling box (z) = 4 (∵ Repetition is not allowed)p = 4 ∴ Total number of 3-digit odd numbers formed= m x n x p = 3 x 5 x 4 = 60. Open in App Suggest Corrections 0 Answer Hint: This is a question based on permutations and combinations. If you see all the alphabets in the word ROSE are different, so we have 4 alphabets. So, actually we need to find the number of permutations of 4 out of these 4 alphabets using the formula $^{4}{{P}_{4}}$ . Complete step-by-step answer: Let us start by interpreting the things given in the question. The word given to us is “ROSE”. It consists of R, O, S, E, i.e., 4 different alphabets and we are asked to find the number of 4-letter words that can be formed using these 4 alphabets. So, basically we need to find the number of permutations of 4 alphabets out of 4 alphabets possible.We know that the permutation of r different objects out of n different objects is given by $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ .Therefore, for the number of permutations of 4 different alphabets, out of the 4 different alphabets of the word LOGARITHM, n=4 and r=4. The number of such permutations is given by $^{n}{{P}_{r}}=\dfrac{4!}{\left( 4-4 \right)!}=\dfrac{4!}{0!}$ .Now we know that the value of 0! Is equal to 1. If we use this value, we get$^{n}{{P}_{r}}=\dfrac{4!}{0!}=\dfrac{24}{1}=24$Hence, the answer to the above question is 24.Note:In questions related to permutations and combinations, students generally face problems in deciding whether the question is based on selection or arrangement. Also, be careful if the repetition of letters is allowed or not, as the answers to both the cases are different. For instance: if the repetition of alphabets was allowed in the above question, the answer would be $4\times 4\times 4\times 4=256$ .
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This video is only available for Teachoo black users Solve all your doubts with Teachoo Black (new monthly pack available now!) Text Solution Solution : Case (i) When the repetition of the letters is not allowed: <br> In this case, the total number of words is the same as the number of ways of filling 4 places `square square square square` by 4 different letters chosen from N, O, S, E. <br> The first place can be filled by any of the 4 letters. Thus, there are 4 ways of filling the first place. Since the repetition of letters is not allowed, so the second place can be filled by any of the remaining 3 letters in 3 different ways. <br> Following it, the third place can be filled by any of the two remaining letters in 2 different ways. And, the fourth palce can be filled by the remaining one letter in 1 way. <br> So, by the fundamental principle of multiplication, the required number of 4 -letter words`=(4xx3xx2xx1)=24`.<br> Hence, required number of words `=24`. <br> Case (ii) When the repetition of the letters is allowed: <br> In this case, each of the four different places can be filled in succession in 4 different ways. <br> So, the required number of 4 -letter words `=(4xx4xx4xx4)=256.` |
How many 4 letter words meaning or without meaning can be formed out of the letters of the word rose when repetition is not allowed?
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