Tips on cracking Aptitude Questions on Combinations

1. TIPS on cracking Aptitude Questions on Combinations

2. Important Formulae • The number of ways of choosing r items from a total of n items, where the order of the chosen items does not matter is denoted by nCror C(n, r) • nCror C(n, r) = P(n, r) = n! • r! r! (n – r)! • C(n, 0)= C(n, n)= 1 [We can choose all available items or none of the items in just one way] • C(n, r)= C(n, n – r) [Choosing r items to be included = choosing (n-r) items for exclusion] • If C(n, x)= C(n, y), then either x= y or x + y = n • C(n, r) + C(n, r – 1) = C(n + 1, r) [Suppose there are n+1 items available and r items need to be selected. The n+1 items can be grouped into two sets: set A with n items and set B with just 1 item. Number of ways of choosing r items from n + 1 items = Number of ways of choosing r items when 1 item in set B is always excluded + Number of ways of choosing r items when 1 item in set B is always included] • Question: If C(n, r)= 84, C(n, r – 1)= 36 and C(n, r+1)= 126, find the value of r. • Solution: • C(n, r) ̳ n – r + 1 ̳ 84 ̳ 7 • C(n, r – 1) r 36 3 • Again, C(n, r + 1) ̳ n – r ̳ 126 ̳ 3 • C(n, r) r + 1 84 2 [Dividing P(n, r) by r! since the order of the chosen r items does not matter] Solving the two equations gives r = 3.

3. Selecting One or More Items When each item is unique: Suppose there are n items and you can choose any number of items. This implies that for each item, there are 2 choices – choose that item or reject it. The total number of choices = 2 * 2 * 2 … n times = 2n Total number of ways of choosing any number of items from a set of n items = nC0 + nC1 + nC2 + nC3 + …. + nCn = 2n Question: There are 7 questions in a question paper. In how many ways can a boy solve one or more questions? Solution: Number of ways in which at least one question can be selected = 27– 1= 127 [Subtracting 1 because the choice of selecting zero questions is not available] When items are not unique: Let there be n things out of which p items are alike of one kind, q items are alike of a second kind, r items are alike of a third kind, and so on.For all p items of the same kind, we can choose 0, 1, 2 .. or p items (Number of choices is p+1). Total number of ways of selecting any number of items = (p+1)(q+1)(r+1)…. Question: In how many ways can you choose the letters of the sentence ‘Daddy did a deadly deed’ ? You can also select no letters at all. Solution: There are 9 D’s, 3 A’s, 3 E’s, 2 Y’s, 1 I and 1 L. Total number of selections= (9+1)(3+1)(3+1)(2+1)(1+1)(1+1) = 1920

4. Dividing items into Groups Suppose we have p + q + r items that are to be divided into 3 groups, first of size p, second of size q and the third of size r. Number of arrangements of all items = (p + q + r)! However, the order within each group does not matter. The number of ways in which p + q + r items can be divided into groups containing p, q and r items respectively = (p + q + r)! p! q! r! Question: In how many ways can 52 playing cards be distributed to 4 players, giving 13 cards to each? Solution: The no. of ways= 52! / (13!)4. p + q + r items ? ? ? ? ? ? ? ? ? ? ? ? Group with q items Group with r items Group with p items

5. Simultaneous Permutations & Combinations Question: How many different words, each containing 2 vowels and 3 consonants, can be formed using all the vowels and 17 consonants? Solution: There are 5 vowels and 17 consonants in all. 2 vowels can be chosen in 5C2 ways and 3 consonants can be chosen in 17C3 ways. Thus, the letters can be selected in 5C2 x17C3 ways. Now, each group of 5 words can be arranged in 5! Ways. Hence, total no. of words= 5C2 x17C3 x 5! = 816000. Question: Out of 3 books on Economics, 4 on Political Science and 5 books on Geography, how many collections can be made if each collection consists of at least one book on each subject? Solution: No. of ways of choosing books on Economics= 23 – 1= 7. No. of ways of choosing books on Political Sciences= 24 – 1= 15. No. of ways of choosing books on Geography= 25 – 1= 31. Therefore, total number of collections that can be formed= 7 x 15 x 31= 3255.

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