Derivations of formulas and their connections

The governing concept is translating a condition into an explicit roster. Decimal digits are 0 through 9, and the strict inequality x < 4 selects 0, 1, 2, and 3. The number 4 is excluded because the condition is not x ≤ 4. Therefore option A is correct. B and C wrongly include 4, while D wrongly omits the digit 0.

Any one of the (n) distinct objects can occupy the first position. In exams think by filling positions one by one.

Since (r) positions are to be filled there are (r) factors. In exams connect the number of factors with the number of positions.

In ordered selection the remaining ((n-r)!) part is removed from (n!). In exams remember ((n-r)!) in the denominator for permutation.

In combination order is not counted so (r!) orders of selected objects are removed. In exams treat combination as selection without order.

First (r) objects are chosen and then arranged in (r!) ways. In exams keep the idea of choose first and arrange later.

In combination extra orders are removed from permutation. In exams write (^{n}P_r) and divide by (r!).

When (r) are chosen the (n-r) objects are automatically left. In exams remember the connection between choosing and leaving.

By (^{n}C_r=^{n}C_{n-r}) we get (k=10-3=7). In exams check the complementary index in equal combinations.

Selecting zero objects is an empty selection and it happens in one way. In exams count empty selection as one way.

The direct answer is option D: there is only one way to choose all n objects. A combination counts selections, not arrangements. If we must choose n objects from a collection containing exactly n objects, every object must be included. There is no freedom of choice, so the result is one selection, written as \(^{n}C_n=1\). Option A is wrong because choosing no object describes \(^{n}C_0=1\), not this expression. Option B is wrong because changing order belongs to permutations; combinations ignore order. Option C is wrong because the objects need not be identical; they are normally distinct. Option D is correct because the complete set itself is the single possible subset. Remember: choosing none or choosing all gives exactly one combination.

The direct answer is option D: one empty way to arrange zero objects. A permutation counts an ordered arrangement of selected objects. If zero objects are selected, there are no positions to fill, and there is exactly one possibility: do nothing, producing the empty arrangement. Algebraically, \({}^{n}P_0=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1\), and this agrees with the convention \(0!=1\). Option A, complete arrangement of \(n\) objects, describes \({}^{n}P_n=n!\), not \({}^{n}P_0\). Option B is false because \(0!\) is valid and equals 1. Option C is false: an empty arrangement is allowed when zero places are being filled. Option D correctly expresses the one empty choice.

The direct answer is option D. The combination formula is \(^{n}C_r=\frac{n!}{r!(n-r)!}\). For \(r=n\), it becomes \(^{n}C_n=\frac{n!}{n!0!}\). Choosing all \(n\) objects has exactly one way, so this expression must equal 1. Since \(n!/n!=1\), the denominator must contain \(0!=1\), giving \(\frac{1}{0!}=1\). Thus the convention \(0!=1\) keeps the boundary case of the combination formula correct. Option A is not the best statement: although \(^{n}C_n=1\), it does not equal \(n!\) in general, so its wording is false. Option B is wrong because \(^{n}P_n=n!\), not 0. Option C is wrong because \(^{n}C_0=1\), not 0. Option D states the exact correct connection. A useful memory cue is that choosing none or choosing everything has one way; the factorial formula must therefore use \(0!=1\).

Arranging all objects gives (n(n-1)\cdots1=n!). In exams connect complete arrangement with factorial.

There are (5) choices for the first place and (4) for the second. In exams choices decrease when repetition is not allowed.

Seating in positions depends on order so permutation is used. In exams when position seat or rank appears check order.

A team has no order so combination is used. In exams remove order for words like group or committee.

First choose (2) objects and then arrange them in (2!) ways. In exams understand permutation as combination times arrangement.

To get combination ordered arrangements are divided by (2!). In exams divide to remove internal orders.

The combination nCr counts selections of r objects from n objects when order does not matter. For r = 1, we are simply choosing one object from n available objects, so there are exactly n possible selections. Algebraically, nC1 = n!/[1!(n−1)!] = n!/(n−1)! = n. Therefore option C is correct. It is not 1, because there are n different single-object choices; it is not n!, because factorial notation counts far more arrangements than required; and it is not 0, because a one-object selection is possible whenever n is at least 1. This identity is also the simplest boundary case of the combination formula.

Only one position is to be filled so there are (n) choices. In exams for (r=1) permutation and combination both equal (n).

In the combination formula after cancelling ((n-2)!) the term (n(n-1)) remains. In exams remember (^{n}C_2) as the pair-making formula.

There are (n) choices for the first place and (n-1) for the second. In exams use the multiplication principle for ordered pairs.

By (^{n}C_r=^{n}C_{n-r}) we get (^{9}C_7=^{9}C_2). In exams use the smaller index to reduce calculation.

Choosing (r) from remaining (n-1) after excluding one fixed object is (^{n-1}C_r). In exams split cases into included and excluded.