Permutation and Combination – Page 16

Question-54860

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Question Number 54860 by Tinkutara last updated on 13/Feb/19 Commented by Tinkutara last updated on 14/Feb/19 $B u t a n s w e r i s$ $\frac{1}{2} (2 n - k^{2} + k - 2)$ Answered by kaivan.ahmadi…

Prove-that-1-n-r-n-1-r-n-1-r-1-2-n-r-n-r-1-n-1-r-3-n-0-n-1-n-2-n-n-2-n-

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Question Number 54740 by gunawan last updated on 10/Feb/19 $P r o v e t h a t$ $1 . (\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n - 1 \\ r \end{matrix}) + (\begin{matrix} n - 1 \\ r - 1 \end{matrix})$ $2 . (\begin{matrix} n \\ r \end{matrix}) + (\begin{matrix} n \\ r - 1 \end{matrix}) = (\begin{matrix} n + 1 \\ r \end{matrix})$ $3 . (\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + . . + (\begin{matrix} n \\ n \end{matrix}) = 2^{n}$ Answered by Kunal12588 last updated on 10/Feb/19…

How-many-words-with-at-least-2-letters-can-be-formed-using-the-letters-from-TINKUTARA-

Tinku Tara June 4, 2023 Permutation and Combination 0 Comments

Question Number 54698 by mr W last updated on 09/Feb/19 $H o w m a n y w o r d s w i t h a t l e a s t 2 l e t t e r s$ $c a n b e f o r m e d u s i n g t h e l e t t e r s f r o m$ $T I N K U T A R A ?$ Answered by afachri last updated on 09/Feb/19 Commented…

Peter-has-12-relatives-5-man-amp-7-woman-and-his-wife-also-has-12-relatives-5-woman-amp-7-man-They-do-not-have-common-relatives-They-decided-to-invite-12-guests-six-each-of-their-relatives-

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Question Number 120209 by benjo_mathlover last updated on 30/Oct/20 $P e t e r h a s 12 r e l a t i v e s (5 m a n & 7 w o m a n)$ $a n d h i s w i f e a l s o h a s 12 r e l a t i v e s$ $(5 w o m a n & 7 m a n) . T h e y d o n o t$ $h a v e c o m m o n r e l a t i v e s . T h e y d e c i d e d$ $t o i n v i t e 12 g u e s t s, s i x e a c h o f$ $t h e i r r e l a t i v e s, s u c h t h a t t h e r e$ $a r e s i x m a n a n d s i x w o m a n$ $${among}\:{the}\:{guests}.\:{How}\:{many} \…

Let-n-be-a-positive-integer-Prove-that-k-0-n-2-k-n-k-n-k-n-k-2-2n-1-n-

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Question Number 120188 by benjo_mathlover last updated on 30/Oct/20 $L e t n b e a p o s i t i v e i n t e g e r .$ $P r o v e t h a t \underset{k = 0}{\sum^{n}} 2^{k} (\begin{matrix} n \\ k \end{matrix}) (\begin{matrix} n - k \\ ⌊ \frac{n - k}{2} ⌋ \end{matrix}) = (\begin{matrix} 2 n + 1 \\ n \end{matrix})$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Such-That-a-n-1-C-r-n-1-n-C-r-n-r-1-b-n-C-0-n-C-2-n-C-4-n-C-1-n-C-3-n-C-5-2-n-1-

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Question Number 54646 by gunawan last updated on 08/Feb/19 $Such That$ $a ._{n + 1} C_{r} = \frac{(n + 1) ._{n} C_{r}}{(n - r + 1)}$ $$\mathrm{b}.\:_{{n}} {C}_{\mathrm{0}} +_{{n}} {C}_{\mathrm{2}} +_{{n}} {C}_{\mathrm{4}…} =_{{n}} {C}_{\mathrm{1}}…

prove-for-r-n-N-k-r-n-k-r-n-1-r-1-Hockey-stick-identity-

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Question Number 185659 by mr W last updated on 25/Jan/23 $p r o v e f o r r, n \in N$ $\underset{k = r}{\sum^{n}} (\begin{matrix} k \\ r \end{matrix}) = (\begin{matrix} n + 1 \\ r + 1 \end{matrix})$ $(H o c k e y - s t i c k i d e n t i t y)$ Commented by mr W last updated…

Question-185642

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Question Number 185642 by Mingma last updated on 24/Jan/23 Commented by mr W last updated on 24/Jan/23 $= C_{4 + 1}^{26 + 1} = C_{5}^{27} = 80730$ Answered…

Find-the-number-of-subsets-of-1-2-3-2000-the-sum-of-whose-elements-is-divisible-by-5-

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Question Number 120090 by bramlexs22 last updated on 29/Oct/20 $F i n d t h e n u m b e r o f s u b s e t s o f$ ${1, 2, 3, \dots, 2000} t h e s u m o f$ $w h o s e e l e m e n t s i s d i v i s i b l e b y 5$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-119896

Tinku Tara June 4, 2023 Permutation and Combination 0 Comments

Question Number 119896 by benjo_mathlover last updated on 27/Oct/20 Commented by bobhans last updated on 28/Oct/20 $(i i) 5 g i r l s m u s t b e t o g e t h e r \Leftrightarrow 5! \times 8!$ Answered by mr W last updated…

Category: Permutation and Combination