![]() ![]() This Permutation and combination formula with an example pdf will help you to know the basics clearly. Permutation and combination sheet pdf is here for effective preparation. So the candidates must know what is permutation and combination and the important permutation and combination formula. Permutation and combination formula pdf is useful for all the bank exams like sbi clerk, sbi po, ibps po, ibps clerk, rrb po, rrb clerk and more. Permutation and combination formula pdf is one of the most effective materials to score well in the quantitative aptitude section of the bank exams and other competitive exams. So you can learn about what is permutation and combination by exploiting this permutation and combination formula for bank exam article or downloading the permutation and combination formula pdf. Aspirants preparing for competitive exams should know what is Permutation and combination. Use this permutation and combination formula pdf for preparation for all bank exams. The aspirants who are effectively preparing for their upcoming bank exams and other competitive exams shall download this permutation and combination formula pdf. This concludes our disucssion on the topic of combination formula.Permutation and Combination Formula PDF for Bank Exam: Permutation and combination formula pdf is available here in this permutation and combination formula for bank exam article. Therefore, the total number of ways = 5C 3 × 12C 5 × 8! = 319334400 ways. These eight letters can arrange themselves in 8! = 8P 8 ways. So, the number of ways of selecting 3 vowels and 5 consonants is 5C 3 × 12C 5. The number of ways of selecting 5 consonants from 12 consonants is 12C 5. Solution: The number of ways of selecting 3 vowels from 5 vowels is 5C 3. Problem: From a word containing 5 vowels and 12 consonants, how many 8 letter words can be formed by using 3 vowels and 5 consonants? ![]() Solution: We know that if nC p = nC q, then either p = q or p + q = n. So, the total number of ways of selections = (a + 1) (b + 1) (c + 1) 2 x – 1. But this includes the case of rejection of all the items. When the items are all different, the number of selecting x items is 2 x. Similarly, the number of ways of selecting b is (b + 1) and so on. Hence there are (a + 1) ways of selecting a. Out of ‘a’ items we can select 0, 1, 2, …, a items. Out of these items, ‘a’ are of the first kind, ‘b’ are of the second kind, ‘c’ are of the third kind and so on and the remaining x are all different. Suppose in a group of n items, there exist some objects which are of a similar kind and a few of them are different. So, the total number of ways of selection of one or more items = 2 n – 1. Therefore, the total number of ways for either situation is 2 × 2 × 2 × 2 … n times = 2 n. Here, each item is categorized in two ways i.e., it may get included or excluded. Let us consider the various situations of selecting r different items from n items when All Items are Different
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